Final answer:
The entropy production for a reversible adiabatic process of an ideal gas is zero. For the actual work input required for adiabatic compression, the formula involving pressures, volumes, and the heat capacity ratio is used, but cannot be calculated without additional information such as the specific volumes.
Explanation:
To determine the amount of entropy produced during an adiabatic compression of air modeled as an ideal gas, we need to recognize that, by definition, an adiabatic process is one in which no heat is transferred to or from the gas. Therefore, assuming the process is also reversible (which it must be, if we are to calculate a non-zero entropy production), the change in entropy (ΔS) for the process would actually be zero. However, as the conditions stated a rise in temperature during the compression, if any irreversibility were present in the real-world scenario, it would indeed generate entropy, but we need more information to calculate the precise amount for a real-world irreversible process.
The minimum theoretical work input for an adiabatic compression can be calculated using the first law of thermodynamics and the relation for adiabatic processes defined as PV^{γ} = constant, where γ (gamma) is the heat capacity ratio (Cp/Cv). For an ideal gas, the work done (W) on the air during adiabatic compression can be expressed using the formula W = (P2*V2 - P1*V1) / (γ - 1), where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. However, since the volume is not given in the question, this calculation cannot be completed without further information.
The entropy produced per kg of air during the adiabatic compression is approximately [tex]\( 0.0348 \)[/tex] kJ/K, and the minimum theoretical work input for the compression is approximately [tex]\( 215.25 \)[/tex] kJ/kg of air.
The amount of entropy produced during the adiabatic compression process in a piston-cylinder assembly, where air is modeled as an ideal gas, can be determined using the following relationship for an adiabatic process:
[tex]\[ s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
where [tex]\( s \)[/tex] is the specific entropy, [tex]\( c_p \)[/tex] is the specific heat at constant pressure, [tex]\( T \)[/tex] is the absolute temperature, [tex]\( R \)[/tex] is the specific gas constant, and [tex]\( P \)[/tex] is the absolute pressure. The subscripts [tex]\( 1 \) and \( 2 \)[/tex] denote the initial and final states, respectively.
Given:
- Initial state: [tex]\( P_1 = 1 \) bar, \( T_1 = 300 \)[/tex] K
- Final state: [tex]\( P_2 = 10 \) bar, \( T_2 = 600 \)[/tex] K
- For air, [tex]\( c_p = 1.005 \) kJ/kg\K and \( R = 0.287 \) kJ/kg[/tex]
First, convert the pressures from bar to Pa (since the gas constant [tex]\( R \)[/tex] is in terms of Pa):
- [tex]\( P_1 = 1 \times 10^5 \) Pa[/tex]
- [tex]\( P_2 = 10 \times 10^5 \) Pa[/tex]
Now, calculate the entropy change:
[tex]\[ s_2 - s_1 = 1.005 \ln\left(\frac{600}{300}\right) - 0.287 \ln\left(\frac{10 \times 10^5}{1 \times 10^5}\right) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \ln(2) - 0.287 \ln(10) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \times 0.693 - 0.287 \times 2.303 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.6965 - 0.6617 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.0348 \text{ kJ/kg\K} \][/tex]
For the minimum theoretical work input during an adiabatic compression, we use the following equation for an ideal gas:
[tex]\[ w_{\text{in,min}} = \int_{V_1}^{V_2} P \, dV \][/tex]
For an adiabatic process, [tex]\( PV^\gamma = \text{constant} \), where \( \gamma \)[/tex] is the heat capacity ratio ([tex]\( c_p/c_v \)[/tex]). The work input can be calculated as:
[tex]\[ w_{\text{in,min}} = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \][/tex]
Using the ideal gas law, [tex]\( PV = mRT \)[/tex], where [tex]\( m \)[/tex] is the mass of the gas, we can express [tex]\( V \)[/tex] in terms of [tex]\( P \) and \( T \)[/tex]:
[tex]\[ V = \frac{mRT}{P} \][/tex]
Since the mass [tex]\( m \)[/tex] cancels out, we can write:
[tex]\[ w_{\text{in,min}} = \frac{mRT_2 - mRT_1}{\gamma - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{R(T_2 - T_1)}{\gamma - 1} \][/tex]
Given that [tex]\( \gamma = \frac{c_p}{c_v} = \frac{c_p}{c_p - R} \)[/tex], we can calculate \( \gamma \) for air:
[tex]\[ \gamma = \frac{1.005}{1.005 - 0.287} \approx 1.4 \][/tex]
Now, calculate the minimum work input:
[tex]\[ w_{\text{in,min}} = \frac{0.287(600 - 300)}{1.4 - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{0.287 \times 300}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{86.1}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} \approx 215.25 \text{ kJ/kg} \][/tex]
Four bricks of length L, identical and uniform, are stacked on top of one another in such a way that part of each extends beyond the one beneath. Find, in terms of L, the maximum values of the following, such that the stack is in equilibrium, on the verge of falling.
To find the maximum values that four bricks of length L can be on the verge of falling while stacked on top of each other in equilibrium, we need to consider the moments of each brick and the conditions for equilibrium. The moments of each brick can be calculated by multiplying the mass of the brick by the distance from the axis of rotation, which is half the length of the brick. By setting the sum of the moments equal to zero, we can find the maximum values for the mass of each brick in terms of L.
Explanation:When four identical and uniform bricks are stacked on top of each other such that part of each brick extends beyond the one beneath, the stack is in equilibrium and on the verge of falling. To find the maximum values in terms of L, we need to consider the moment of each brick and the conditions for equilibrium. The moment is the product of the mass of the brick and the distance from the axis of rotation, which is half the length of the brick.
Let's assume the bricks have a length L and that they are stacked vertically with their centers of mass aligned. The distance from the axis of rotation to the center of mass is L/2 for each brick. The moment of the top brick is then (mass of the brick) * (L/2). The moment of the second brick is (mass of the brick) * (3L/2), considering the length of the first brick as the distance from the axis of rotation. Similarly, the moment of the third brick is (mass of the brick) * (5L/2), and the moment of the fourth brick is (mass of the brick) * (7L/2).
For the stack to be in equilibrium, the sum of the moments must be zero. Therefore, we have the equation:
(mass of the first brick) * (L/2) + (mass of the second brick) * (3L/2) + (mass of the third brick) * (5L/2) + (mass of the fourth brick) * (7L/2)
By simplifying this equation and substituting the mass of each brick with a constant value (let's say M), we can find the maximum values for the mass of each brick such that the stack is in equilibrium on the verge of falling, given a certain length L.
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With the assumption of no slipping, determine the mass m of the block which must be placed on the top of the 6.5-kg cart in order that the system period be 0.66 s. What is the minimum coefficient of static friction for which the block will not slip relative to the cart if the cart is displaced 67 mm from the equilibrium position and released?
Answer:
The minimum coefficient of static friction should be 0.62.
Explanation:
Given that,
Mass of block = m
Mass of cart = 6.5 kg
Time period = 0.66 s
Displacement = 67 mm
We need to calculate the mass of block
Using formula of time period
[tex]T=2\pi\times(\dfrac{m}{k})[/tex]
Put the value into the formula
[tex]0.66=2\pi\times(\dfrac{m+6}{600})[/tex]
[tex]m=\dfrac{0.66\times600}{4\pi^2}-6[/tex]
[tex]m=4.03\ kg[/tex]
We need to calculate the maximum acceleration of SHM
Using formula of acceleration
[tex]a_{max}=\omega^2 A[/tex]
Maximum force on mass 'm' is [tex]m\omega^2 A[/tex]
Which is being provided by the force of friction between the mass and the cart.
[tex]\mu_{s}mg \geq m\omega^2 A[/tex]
[tex]\mu_{s}\geq \dfrac{\omega^2 A}{g}[/tex]
[tex]\mu_{s} \geq (\dfrac{2\pi}{T})^2\times\dfrac{A}{g}[/tex]
Put the value into the formula
[tex]\mu_{s} \geq (\dfrac{2\pi}{0.66})^2\times\dfrac{0.067}{9.8}[/tex]
[tex]\mu_{s} \geq 0.62[/tex]
Hence, The minimum coefficient of static friction should be 0.62.
The deepest point in any ocean is in the Mariana Trench, which is about 11 km deep, in the Pacific. The pressure at this depth is huge, about 1.13 108 N/m2. (Take the bulk modulus of seawater to be 2.34 109 N/m2) (a) Calculate the change in volume of 0.9 m3 of seawater carried from the surface to this deepest point. (b) The density of seawater at the surface is 1.03 103 kg/m3. Find its density at the bottom
Answer:
a) ΔV = - 4.346 10⁻² , b) ρ’= 1.082 10³ kg / m³
Explanation:
The volume module is defined as the ratio of the pressure and the unit deformation, with a negative sign, for the module to be positive
B = - P / (ΔV/V)
a) The ΔV volume change
ΔV/V = -P / B
ΔV = - P V / B
ΔV = - 1.13 10⁸ 0.9 /2.34 10⁹
ΔV = - 4.346 10⁻²
b) Density at the bottom of the sea
On the surface
ρ = m / V
m = ρ V
m = 1.03 10³ 0.9
m = 0.927 10³ kg
Body mass does not change with depth
Deep down
ρ’= m / V’
ΔV = 4.346 10⁻²
[tex]V_{f}[/tex]- V₀ = 4,346 10⁻²
[tex]V_{f}[/tex] = 0.0436 + Vo
[tex]V_{f}[/tex]= -0.04346 + 0.9
[tex]V_{f}[/tex] = 0.85654 m³
ρ’= 0.927 10³ / 0.85654
ρ’= 1.082 10³ kg / m³
Using the formula for volume change under pressure and density calculations, it can be determined that the change in volume of 0.9 m³ seawater when taken to the Mariana Trench is -0.043 m³ and its density at the bottom is approximately 1072 kg/m³.
Explanation:The pressure in the ocean increases with depth due to the weight of the overlying water. This high pressure can compress the volume of the water at such depths, altering its density. As the question provides, the bulk modulus of seawater is given as 2.34 x 10⁹ N/m² and the pressure in the Mariana Trench is 1.13 x 10⁸ N/m².
(a) To find the change in volume, we can use the formula ΔV = -(PΔV/B), where P is the pressure, ΔV is the change in volume, and B is the bulk modulus. Inserting the given values, we get the change in volume to be -0.043 m³.
(b) The density of a substance is its mass divided by its volume. At the bottom of the Mariana Trench, the volume of water has decreased due to the high pressure, but its mass remains the same. Therefore, as the volume decreases, the density increases. The new density, ρ', is calculated using the formula ρ' = ρ/(1-ΔV/V), where ρ is the initial density and V is the initial volume. Substituting into this equation, we get ρ' = 1.03 x 10³ kg/m³ / (1 -(-0.043), which gives us a density at the bottom of the Mariana Trench of approximately 1072 kg/m³.
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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1 = 12kg and diameter d1 = 60cm is at rest (f1 = 0min−1 ) and disengaged from disk 2 with mass m2 = 8kg and diameter d2 = 40cm that is rotating with a frequency of f2 = 200min−1 . When we engage the clutch both disks become connected and disk 1 is accelerated while disk 2 decelerates due to a portion of its rotational energy being used to accelerate disk 1. In full contact both disks are rotating with the same angular velocity. Calculate this final angular velocity ω and the corresponding frequency f. We are neglecting further loss of energy due to heat as a result of friction. This friction and heat is the reason why real clutches wear out over time. The moment of inertia for a solid disk can be found in the textbook.
Answer:
w = 4,786 rad / s , f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
[tex]L_{f}[/tex] = (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ = [tex]L_{f}[/tex]
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz
A coil of wire 0.12 m long and having 340 turns carries a current of 13 A. (a) What is the magnitude of the magnetic field strength H (in A/m)? (b) Calculate the flux density B (in tesla) if the coil is in a vacuum. (c) Calculate the flux density (in tesla) inside a bar of metal positioned within the coil that has a magnetic susceptibility of 1.90 x 10-4. (d) Calculate the magnitude of the magnetization M (in A/m).
To solve this problem it is necessary to apply the concepts related to the magnetic field
the flux density and the magnitude of the magnetization.
Each of these will be tackled as the exercise is carried out, for example for the first part we have to:
Part A) Magnitude of the magnetic field
[tex]H = \frac{NI}{L}[/tex]
Where,
N = Number of loops
I = Current
L = Length
If we replace the given values the value of the magnitude of the magnetic field would be:
[tex]H= \frac{340*13}{0.12}[/tex]
[tex]H = 36833 A\cdot turns/m[/tex]
For the second and third part we will apply the concepts of density both in vacuum and positioned at a point, like this:
PARTE B) Flux density in a vacuum
[tex]B = \mu_0 H[/tex]
Where,
[tex]\mu_0 =[/tex] Permeability constant
[tex]B = (4\pi*10^{-7})(36833)[/tex]
[tex]B = 0.04628T[/tex]
PART C) To find the Flux density inside a bar of metal but the magnetic susceptibility is given
[tex]X_m = 1.9*10^{-4}[/tex]
[tex]\mu_R = 1+X_m[/tex]
[tex]\mu_R = 1.00019[/tex]
Then the flux density would be
[tex]B = \mu_0 \mu_R H[/tex]
[tex]B = (4\pi*10^{-7})(1.00019)(36833)[/tex]
[tex]B = 0.04629T[/tex]
PART D) Finally, the magnetization describes the amount of current per meter, and is given by the magnetic susceptibility, that is:
[tex]M = X_m H[/tex]
[tex]M = 1.9*10^{-4}*36833[/tex]
[tex]M = 6.996A/m[/tex]
This solution calculates the magnetic field strength, flux density and magnetization in a wire coil, as well as the flux density inside a metal bar within the coil, using Ampere's law and the formulas for magnetic field and flux densities.
Explanation:The calculations are derived from Ampere's law and the formulas for magnetic field strength and flux density. For part (a), to find the magnitude of the magnetic field strength H (in A/m), we consider the given length of the coil and the number of turns. H= nI, where n is the number of turns per unit length and I is the current. In the given case, n= 340/0.12 = 2833.33, so H = 2833.33 turns/m * 13 A = 36833 A/m.
For part (b), we need to calculate the flux density B (in T) inside the coil which is in a vacuum. Using the formula B = μH, where μ = 4π x 10-7 T. m/A is the permeability of vacuum, we find B= 4π x 10^-7 T.m/A * 36833 A/m = 4.6 x 10^-2 T.
For part (c), we need to calculate the flux density B inside a bar of metal with susceptibility χ_m = 1.90 x 10-4. In this case, B = μ(H + M), where M = χ_m H is the magnetization. We find M= 1.90 x 10^-4 * 36833 A/m = 7 A/m, so B in metal = 4π x 10^-7 T.m/A * (36833 A/m + 7 A/m) = 4.6 x 10^-2 T.
For part (d), we've already calculated the magnitude of the magnetization M to be 7 A/m.
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Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions, the wheel comes to rest on a space that has a $1500 value prize. If the initial angular speed of the wheel is 3.50 rad/s, find the angle through which the wheel has turned when the angular speed reaches 2.00 rad/s.
Answer:
Explanation:
Given
after 5.5 revolution wheel comes to stop
i.e. radian turned before stopping
[tex]\theta =2\pi \times 5.5 rad[/tex]
initial angular velocity [tex]\omega _0=3.5 rad/s[/tex]
[tex]\omega ^2-\omega _0^2=2\cdot \alpha \cdot \theta [/tex]
where [tex]\alpha =angular\ acceleration\ or\ deceleration[/tex]
[tex]0-(3.5)^2=2(\alpha )(2\pi \cdot 5.5)[/tex]
[tex]\alpha =-0.1772 rad/s^2[/tex]
angle turned when final angular velocity is [tex]2 rad/s[/tex]
[tex]2^2-3.5^2=2\cdot (-0.1772)(\theta )[/tex]
[tex]\theta =23.27\ radians[/tex]
A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the speaker? What is the wavelength of this sound wave as it travels through air with a speed of about 341 m/s? What is the wavelength of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?
Answer:
0.003034 s
1.035 m
4.5 m
Explanation:
[tex]f[/tex] = frequency of the tone = 329.6 Hz
[tex]T[/tex] = Time period of the sound wave
we know that, Time period and frequency are related as
[tex]T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s[/tex]
[tex]v[/tex] = speed of the sound in the air = 341 ms⁻¹
wavelength of the sound is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m[/tex]
[tex]v[/tex] = speed of the sound in the water = 1480 ms⁻¹
wavelength of the sound in water is given as
[tex]\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m[/tex]
A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque, and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot. The 2.60-N force is at an angle of 30.0o to the rod and causes a cw torque. What is the net torque about the pivot?
From the definition we have that the Torque corresponds to the Multiplication between the Force (or its respective component) and the radius of distance of the Force to the inertial turning point.
Mathematically this can be expressed,
[tex]\tau = F \times d[/tex]
Where,
F = Perpendicular component of force
d = distance from pivot point
The total sum of the torques would be equivalent to
[tex]\tau_{net} = \tau_1 +\tau_2[/tex]
According to the values given, torque 1 and 2 would be given by
[tex]\tau_1 = 6*1.2 = 7.2N\cdot m (+)[/tex]
[tex]\tau_2 = -5.2sin(30) = -7.8N\cdot m (-)[/tex]
Therefore the net Torque is
[tex]\tau_{net} = \tau_1+\tau_2[/tex]
[tex]\tau_{net} = 7.2-7.8[/tex]
[tex]\tau_{net} = -0.6N\cdot m[/tex]
Therefore the net torque about the pivot is -0.6Nm
An adventurous stdent stands on top of a spinning chair holding a Jack-o-lantern in each hand (each of which is 5.0 kg) as she extents her arms horizontally.
She spins around, making one revolution in 2.0 s.
The moment of inertia when her arms are outstretched is 13.0 kgm, and if she brings the Jack-o-lanterns in close to her stomach, the moment of inertia drops to 2.6 kgm?.
What is the final angular velocity?
(A) 2.5 rev/s
(B) 2.5 rad/s
(C) 27 rad/s
(D) 0.5 rev/s
(E) 5.0 rev/s
Answer:
option A
Explanation:
the initial moment of inertia of system , I₀ = 13 kg.m²
the final moment of inertia of the system , I = 2.6 kg.m²
time = t = 2 s
mass = 5 Kg
the initial angular speed ,
[tex]\omega = \dfrac{2\pi}{T}[/tex]
[tex]\omega = \dfrac{2\pi}{2}[/tex]
ω = 3.14 rad/s
[tex]\omega = \dfrac{3.14}{2\pi}[/tex]
[tex]\omega_0= 0.5\ rev/s[/tex]
let the final angular speed be ω₀
using conservation of angular momentum
I₀ x ω₀= I x ω
13 x ω₀= I x ω
13 x 0.5 = 2.6 x ω
ω = 2.5 rev/s
hence, the correct answer is option A
A 28 kg child goes down a straight slide inclined 38∘ above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30 N. How large is the normal force of the slide on the child? If the child slides with a constant speed, how large is the kinetic frictional force from the slide?
The normal force of the slide on the child is 217.3 N and the kinetic frictional force from the slide is 166.8 N when the child slides down the slide at a constant speed.
Explanation:The normal force on a slope, which is always perpendicular to the surface, is equal to the weight component of the object that is perpendicular to the slope. As the child slides down the slide at a constant speed, the net force on the child is zero. In this scenario, let's denote mass (m) as 28 kg, inclination angle (θ) as 38 degrees, and g as gravitational acceleration which is 9.8 m/s². So, the normal force (N), which is equal to m*g*cosθ, can be calculated as: 28 kg * 9.8 m/s² * cos(38) = 217.3 N.
The frictional force from the slide acts in the opposite direction to the motion. When the sliding speed is constant, this kinetic frictional force equals the component of the child's weight that is parallel to the slope (m*g*sinθ). Hence, the kinetic frictional force would be: 28 kg * 9.8 m/s² * sin(38) = 166.8 N.
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The normal force of the slide on the child can be found by multiplying the child's weight by the cosine of the angle of inclination. The kinetic frictional force from the slide is equal to the horizontal force exerted by the rope.
Explanation:To find the normal force of the slide on the child, we need to determine the component of the child's weight perpendicular to the slide. Since the slide is inclined at 38∘ above horizontal, the normal force is equal in magnitude to the component of the child's weight perpendicular to the slide, which is given by:
Normal force = weight * cos(38∘)
Next, to find the kinetic frictional force from the slide, we need to use the horizontal force exerted by the rope. Since the child slides with a constant speed, the kinetic frictional force must be equal in magnitude to the horizontal force exerted by the rope, which is given as 30 N.
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The buildup of plaque on the walls of an artery may decrease its diameter from 1.1 cm to 0.90 cm. The speed of blood flow was 17 cm/s before reaching the region of plaque buildup.
A). Find the speed of blood flow within the plaque region. This was 25 cm/s.
B.) Find the pressure drop within the plaque region. I got 17.808 but says I'm wrong. Am I missing a step?
Answer
Initial radius of the artery is (1.1 cm) / 2 = 0.55 cm
final radius of the artery is (0.90 cm) / 2 = 0.45 cm
initial velocity of the blood is 17 cm/s
Using equation of continuity is
A₁v₁=A₂v₂
π r₁² x v₁ = π r₂² x v₂
r₁² x v₁ = r₂² x v₂
0.55² x 17 =0.45² x v₂
v₂=25.39 cm/s
Bernoulli's equation is
[tex]P_1 - P_2 = \dfrac{1}{2}\rho (v_2^2-v_1^2)[/tex]
rho is the density of blood = 1060 kg/m^3
[tex]P_1 - P_2 = \dfrac{1}{2}\times 1060 \times (0.254^2-0.17^2)[/tex]
[tex]P_1 - P_2 =18.87\ Pa[/tex]
The student uses the continuity equation to correctly solve the first part, but for pressure drop, they need to apply Bernoulli's principle. Using the equation P1 + 1/2ρv1² = P2 + 1/2ρv2², with ρ being the blood's density and v1 and v2 the speeds, they can find the pressure drop. However, real-life complications due to blood viscosity and turbulence might affect the results.
Explanation:The subject of this question is Physics (specifically, fluid dynamics). The concept being applied here is that of continuity and Bernoulli's principle. The continuity equation for fluid states that the mass flow rate must be constant throughout the length of the pipe. This translates to: Area 1 × Speed 1 = Area 2 × Speed 2. You correctly set up this equation to find Speed 2, which is the speed of the blood flow within the plaque region.
For the pressure drop, we need to use Bernoulli's principle, which describes that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant in a non-viscous, steady-flowing system. Applying Bernoulli's Equation: P1 + 1/2ρv1² = P2 + 1/2ρv2². Solving this equation gives us the pressure difference within the plaque region.
Keep in mind that you might have to use the change in blood speed, and also the density of blood within the equations to find the accurate pressure drop. Also, Bernoulli's principle applies to ideal situations and there could be changes in real-life situations due to the viscosity of blood and turbulence caused by plaque.
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For years, marine scientist were mystified by sound waves detected by underwater microphones in the Pacific Ocean. These so-called T waves were among the purest sounds in nature. Eventually the researchers traced the source to underwater volcanoes whose rising columns of bubbles resonated like organ pipes. A typical T wave has a frequency of 6.8 Hz. Knowing that the speed of sound in seawaver is 1,530 m/s, determine the wavelength of a T wave.
Answer:
225 m
Explanation:
[tex]f[/tex] = Frequency of the T wave = 6.8 Hz
[tex]v[/tex] = Speed of sound in seawater = 1530 ms⁻¹
[tex]\lambda[/tex] = Wavelength of the T wave
we know that, frequency, speed and wavelength are related as
[tex]wavelength = \frac{speed}{frequency}[/tex]
[tex]\lambda = \frac{v}{f}[/tex]
Inserting the values, we get
[tex]\lambda = \frac{1530}{6.8}\\\lambda = 225 m[/tex]
Final answer:
The wavelength of a T wave with a frequency of 6.8 Hz in seawater, where the speed of sound is 1,530 m/s, is calculated to be approximately 225 meters using the formula for wave speed.
Explanation:
To calculate the wavelength of a T wave with a frequency of 6.8 Hz in the Pacific Ocean, where the speed of sound in seawater is 1,530 m/s, we can use the formula for wave speed: v = f × λ, where v is the speed of sound, f is the frequency, and λ is the wavelength.
By rearranging the formula to solve for the wavelength (λ = v / f), and substituting in the known values, we get λ = 1,530 m/s / 6.8 Hz, which calculates to approximately 225 meters. Therefore, the wavelength of a T wave with a 6.8 Hz frequency in seawater is about 225 meters.
Each milligram of glucose has the same amount of energy available to do work. The series B test tubes produced more bacteria per milligram of glucose than did the series A test tubes. Assuming that each bacterium produced requires a certain amount of energy, which test tube should contain some products of glucose that still contain some "unused" energy?
Answer:
The series A test tube has some left amount of glucose left in it.
Explanation:
Let's assume that a fixed amount of glucose is synthesized, for the fixed quantity the bacteria produced in A and B be x and y respectively,
Therefore, the condition on x and y is, y > x as the no. of bacteria present in B is greater.
As a result B would require a greater amount of energy for its functioning, these energy would be derived from the already fixed amount of glucose present.
A test tube would also require the energy for its x number of bacteria, but it is less than that of B.
Therefore, there would be some unused glucose left in Test Tube Series A which has unused energy.
A ball of mass 0.800 kg is carefully balanced on a shelf that is 3.00 m above the ground. What is its gravitational potential energy?a. 2.08J b. 20.4J c. 7.84J d. 25.5J
Potential energy is the mechanical energy associated with the location of a body within a force field (gravitational, electrostatic, etc.) or the existence of a force field inside a body (elastic energy)
The gravitational potential force is subject to the relationship
[tex]PE = mgh[/tex]
Where,
m = mass
g = Gravitational Energy
h = Height
If our height is 3 meters, the mass is 0.8Kg and the earth has a gravitational acceleration of 9.8m / s, we will have to
PE = (9.8)(3)(0.8)
PE = 23.52J
The closest answer is D.
Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
Answer:
Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.
In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.
Final answer:
Elastic collisions conserve both momentum and kinetic energy with objects bouncing off each other, while inelastic collisions conserve momentum but not kinetic energy, often resulting in the objects sticking together or showing permanent deformations. Mathematically analyzing an elastic collision is generally easier due to the conservation of both momentum and kinetic energy.
Explanation:
The primary physical difference between elastic and inelastic collisions lies in how kinetic energy is conserved in the collision process. In an elastic collision, both momentum and kinetic energy are conserved. This means that after the collision, the total kinetic energy of the two objects remains the same as before they collided. The objects bounce off each other and do not stick together. As a helpful trick, remember that 'elastic' implies the ability to return to the original shape, just like objects in an elastic collision bounce off one another and return to their separate ways.
On the other hand, an inelastic collision is one where the kinetic energy is not conserved, although the momentum is still conserved. The objects may stick together after the collision, which can be referred to as a perfectly inelastic collision. Since kinetic energy is lost in inelastic collisions, the objects typically do not separate after colliding; rather, they may move as a single entity or with less kinetic energy than they had initially. An inelastic collision involves a permanent deformation of the colliding bodies or the generation of heat.
To your other questions: Analyzing an elastic collision might be easier, as both momentum and kinetic energy conservation can be used to solve for final velocities, while in an inelastic collision, only momentum is conserved. Verification can be done both mathematically and graphically, but mathematical methods often provide a direct computation of the final velocities. When considering the recoil in two objects thrown by two people, if the objects have the same velocity but different masses, question 15 from 8.3 suggests that the person who threw the heavier object will gain more velocity upon recoil. However, in reality, since momentum is conserved, both individuals would experience the same amount of recoil velocity, opposite in direction to the thrown object.
A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0.54 m below the surface of the water. The water surface drops very slowly and its speed is approximately zero. Keep 2 decimal places in all answers. (a) Find the speed v (in m/s) at which water squirts out of the hole.
Speed can be found through the application of concepts related to potential energy and kinetic energy, for which you have
[tex]KE = PE[/tex]
[tex]\frac{1}{2}mv^2 = mgh[/tex]
Where,
m = mass
v = Velocity
g = Gravitational acceleration
h = Height
Re-arrange to find the velocity we have,
[tex]v^2 = 2gh[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.8)(0.54)}[/tex]
[tex]v = 3.253m/s[/tex]
Therefore the speed at which water squirts out of the hole is .3253m/s
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω , and the reactance of the capacitor is 473 Ω . The voltage amplitude across the capacitor is 364 V . Part A What is the current amplitude in the circuit? II = nothing A Request Answer Part B What is the impedance? ZZ = nothing Ω Request Answer Part C What two values can the reactance of the inductor have? Enter your answers in ascending order separated by a comma. XLX L = nothing Ω Request Answer Part D For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? the less value of XL the larger value of XL Request Answer Provide Feedback
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
The current amplitude in the circuit is 1.51A. The impedance of the circuit is 168Ω. The reactance of the inductor can have two values, which can be calculated using the equation XL = 2&pivL.
Explanation:The current amplitude in the circuit can be calculated using the equation:
II = Vo/R = 116V / 77.0Ω = 1.51A.
The impedance of the circuit can be calculated using the equation:
ZZ = √(R2 + (XL - XC)2) = √(77.0Ω2 + (473Ω - 364Ω)2) = 168Ω.
The reactance of the inductor can have two values. From the equation XL = 2&pivL, we can set XL = XC = 473Ω and solve for L, which gives L = XC / (2&piv) = 473Ω / (2πf) where f is the frequency of the AC voltage source.
To find the values for which the angular frequency is less than the resonance angular frequency, we need to compare the values of XL obtained for different frequencies. If the frequency is such that XL > XC, then the angular frequency is less than the resonance angular frequency.
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For a fiber-reinforced composite, the efficiency of reinforcement η is dependent on fiber length l according to where x represents the length of the fiber at each end that does not contribute to the load transfer. What length is required for a 0.62 efficiency of reinforcement, assuming that x is 0.61 mm?
Answer:
l = 3.21 mm
Explanation:
The efficiency of a fiber-reinforced is the following:
[tex] \eta = \frac{l - 2x}{l} [/tex]
Where:
η: is the efficiency
l: is the fiber length
x: is the length of the fiber at each end that doesn't contribute to the load transfer
So the length required for a 0.62 efficiency of reinforcement is:
[tex] l = \frac{2x}{1- \eta} = \frac{2 \cdot 0.61 mm}{1- 0.62} = 3.21 mm [/tex]
I hope it helps you!
The length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite can be calculated using the equation η = (l-x)/l, where l represents the total length of the fiber and x represents the length at each end that does not contribute to load transfer. By rearranging the equation, we can solve for l.
Explanation:To calculate the length required for a 0.62 efficiency of reinforcement in a fiber-reinforced composite, we can use the equation η = (l-x)/l. Here, l represents the total length of the fiber, and x represents the length at each end that does not contribute to load transfer. We are given that x is 0.61 mm. We can rearrange the equation to solve for l:
η = (l-x)/l
0.62 = (l-0.61)/l
0.62l = l - 0.61
0.62l - l = -0.61
0.62l = -0.61
l = -0.61/0.62
l = -0.983
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Calculate the number of molecules in a deep breath of air whose volume is 2.35 L at body temperature, 37 ∘C, and a pressure of 735 torr. Express the answer in molecules to three significant figures. NN = nothing molecules Request Answer Part B The adult blue whale has a lung capacity of 5.0×103 L. Calculate the mass of air (assume an average molar mass 28.98 g/mol) contained in an adult blue whale’s lungs at 0.5 ∘C and 1.07 atm, assuming the air behaves ideally. Express the answer in kilograms to two significant figures. mm = nothing kg
Answer:
Part A. 5.36x10²³ molecules of air
Part B. 6.9kg
Explanation:
Part A.
To calculate the number of molecules of the air we need first find the number of moles of air using the equation of ideal gas law:
[tex] PV = nRT [/tex] (1)
where P: is the pressure, V: is the volume, n: is the number of moles of the gas, R: is the gas constant and T: is the temperature
[tex] n = \frac{PV}{RT} = \frac{735torr \cdot 1atm/760torr \cdot 2.35L}{0.082 Latm/Kmol \cdot (37 + 273)K} = 0.089 moles [/tex]
Now by using the Avogadro's number we can find the number of molecules of air:
[tex] number of molecules = \frac{6.022 \dot 10^{23}}{1mol} \cdot 0.089moles = 5.36 \cdot 10^{22} molecules [/tex]
Part B.
Similarly, to calculate the mass of air first we need to detemine the number of moles using equation (1):
[tex] n = \frac{PV}{RT} = \frac{1.07atm \cdot 5.0\cdot 10^{3}L}{0.082 Latm/Kmol \cdot (0.5 + 273)K} = 238.55 moles [/tex]
So, the mass of air is:
[tex] m = moles \cdot M [/tex]
where M: is the average molar mass of air
[tex] m = 238.55moles \cdot 28.98g/mol = 6.9 kg [/tex]
I hope it helps you!
The student needs to use the ideal gas law and Avogadro's number to convert the volume of air to moles and then to molecules. For the blue whale, use the ideal gas law to convert volume to moles, then multiply by the molar mass of air to get mass.
Explanation:To begin with, we first convert the volume into molecules. Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is the number of moles, R is the gas constant and T is temperature. The number of moles, n, can be found by rearranging the equation to n=PV/RT.
Substituting the given values (converting pressure to atm and volume to L, and temperature to Kelvin), we will compute n. After finding the no of moles, the number of molecules is calculated by multiplying the number of moles by Avogadro's number.
In the case of the blue whale, we again use the ideal gas law to find the number of moles of air in the lungs, then multiply by the molar mass of air (28.98 g/mol) to find the mass in grams. Finally, we then convert from grams to kilograms.
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A(n) ____ is a solar system object that enters Earth's atmosphere and becomes very hot due to friction between the object and Earth's atmosphere and does not survive to reach the surface.
a. asteroid
b. meteor
c. comet
d. meteoroid
e. planetesimal
Answer:
d. meteoroid
Explanation:
An asteroid is a small rocky mass of substance that orbits around the sun. They are smaller than a planet in size but larger than a pebble sized mass called meteoroids.Sometimes a meteoroid comes close enough to the Earth and enters the Earth’s atmosphere, due to friction with the atmosphere it vaporizes and turns into a meteor appearing as a streak of light in the sky called meteor.Comets are the mass of ice and dust revolving around the sun. They keep coming closer to the sun in a helical path and when there is enough heat these get vapourised leaving behind a tail of vapour and dust.When one asteroid smashes into another breaking off small pieces which are called meteoroids if they get completely burn into ashes or vapourize when passing through the atmosphere of the earth, but if they fall as a mass of rock they are known as meteorites.A planetesimal is a solid object that arises due to the accumulation of orbiting bodies whose internal strength is dominated by self-gravity and whose orbital dynamics is not significantly affected by gas drag.Celsius
The world's most common temperature scale is Celsius. Abbreviated C, it is virtually the same as the old centigrade scale and therefore has 100 degrees between the melting point and boiling point of water, taken to occur at 0 and 100 degrees, respectively.
Kelvin
Temperature is a measure of the thermal energy of a system. Thus cooling can proceed only to the point at which all of the thermal energy is removed from the system, and this process defines the temperature of absolute zero. The Kelvin scale, also called the absolute temerature scale, takes its zero to be absolute zero. It uses units of kelvins (abbreviated K), which are the same size as the degrees on the Celsius scale.
Fahrenheit This anachronistic temperature scale, used primarily in the United States, has zero defined as the lowest temperature that can be reached with ice and salt, and 100 degrees as the hottest daytime temperature observed in Italy by Torricelli.
Required:
A. In the equation of state for the perfect gas, pV = nRT , which of the following three temperature scales must be used?
O Celsius
O Kelvin
O Fahrenheit
Answer:
In this equation the scale used must be Kelvin
Explanation:
The absolute temperature is the value of the measured temperature relative to a scale starting at Absolute Zero (0 K or -273.15 °C). It is one of the main parameters used in thermodynamics and statistical mechanics. In the international system of units it is expressed in kelvin, whose symbol is K
The absolute temperature should always be used in the ideal gas state equation as this can only be in kelvin grade scale.
In grinding a steel knife blade (specific heat = 0.11 cal/g-c),the metal can get as hot as 400C. If the blade has a mass of 80g,what is the minimum amount of water needed at 20C if the water isnot to rise above the boiling point when the hot blade is quenchedin it?
Answer:
33 g.
Explanation:
Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.
As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.
Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.
So, we can write the following equation:
cs*mk*Δtk = cw*mw*Δtw
Replacing by the givens of the question, we have:
0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC
Solving for mw = 2,640 cal / 80 cal/g =33 g.
If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I find half of the atmosphere air pressure?
Answer: 5.5km
Explanation:
Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).
Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.
Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.
It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).
For a block to move down an inclined plane what force has to be the greatest?
A. Compression
B. Normal
C. tension
D. Gravity
E. Shear
Answer:
D) True. This is what creates the body weight
Explanation:
Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)
X axis
Wx -fr = ma
Y Axis
N - Wy = 0
With trigonometry we can find the components of weight
sin θ = Wₓ / W
cos θ = [tex]W_{y}[/tex] / W
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
W sin θ - fr = ma
From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body
Let's examine the answers
A False This force does not apply because it is not a spring
B) False. It is balanced at all times with the component (Wy) of the weight
C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower
D) True. This is what creates the body weight
E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points
For a block to move down an inclined plane, the force of gravity must be greater than the other forces. This includes the normal force and any friction that may be present. Other forces such as compression, tension, and shear are not directly involved in this process.
Explanation:For a block to move down an inclined plane, the greatest force must be gravity. The force of gravity acts downward and causes the block to slide down the slope. This force must be stronger than the others such as the normal force (the force exerted by the plane on the block) and any friction forces that may be present.
The other forces listed (compression, tension, and shear) are not directly related to the movement of the block down the inclined plane. Compression and tension are forces that act in opposite directions either pushing (compression) or pulling (tension) an object. Shear is a force that causes materials to slide past each other and is not directly applicable to this scenario
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Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you fall out of your boat and immediately grab a piling of the Warm Springs Bridge. You hold on for 40 s and then swim after the boat with a speed relative to the water of 0.95 m/s. The distance of the boat downstream from the bridge when you catch it is______________.
Answer:
d = 142.5 m
Explanation:
This is a vector exercise. Let's calculate how much the boat travels in the 40s
d₀ = [tex]v_{b}[/tex] t
d₀ = 0.75 40
d₀ = 30 m
Let's write the kinematic equations
Boat
x = d₀ + [tex]v_{b}[/tex] t
x = 0 + [tex]v_{h}[/tex] t
At the meeting point the coordinate is the same for both
d₀ + [tex]v_{b}[/tex] t = [tex]v_{h}[/tex] t
t ( [tex]v_{h}[/tex] - [tex]v_{b}[/tex]) = d₀
t = d₀ / ( [tex]v_{b}[/tex]- [tex]v_{h}[/tex])
The two go in the same direction therefore the speeds have the same sign
t = 30 / (0.95-0.775)
t = 150 s
The distance traveled by man is
d = [tex]v_{h}[/tex] t
d = 0.95 150
d = 142.5 m
The distance of the boat downstream when you catch it is 60 meters.
Explanation:To find the distance of the boat downstream when you catch it, we can use the equation d = vt, where d is the distance, v is the velocity, and t is the time.
Given that the current of the river is 0.75 m/s and you hold onto the piling for 40 seconds, the distance drift with the current is:
ddrift = (0.75 m/s)(40 s) = 30 mAfter you start swimming with a speed of 0.95 m/s and catch up to the boat, the distance you swim against the current is equal to the distance the boat drifts:
dswim = ddrift = 30 mTherefore, the total distance downstream when you catch the boat is:
d = ddrift + dswim = 30 m + 30 m = 60 mHence, the total distance downstream when you catch the boat is 60 m.
An 800-kHz radio signal is detected at a point 8.5 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 0.90 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the average electromagnetic energy density at that point? (c = 3.0 x 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
A. 7.2 pJ/m3
B. 10 pJ/m3
C. 3.6 pJ/m3
D. 14 pJ/m3
E. 5.1 pJ/m3
To solve this problem we need to apply the concepts related to the average electromagnetic energy density. Which is given as
[tex]U = \frac{1}{2}\epsilon_0 E^2[/tex]
Where,
\epsilon_0 = Permettivity of free space constant
E = Electric Field amplitude
Since the average electromagnetic energy density is directly proportional to the amplitude of the magnetic field then we have to
[tex]E = \frac{1}{2} (8.85*10^{-12}C^2/N\cdot m^2)(0.9V/m)^2[/tex]
[tex]E = 3.6*10^{-12}J/m^3[/tex]
[tex]E = 3.6pJ/m^3[/tex]
Therefore the correct answer is C.
A rocket is fired vertically upward. At the instant it reaches an altitude of 2700 m and a speed of 274 m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 235 m/s following the explosion. The second fragment has a speed of 484 m/s and is moving east right after the explosion. What is the magnitude of the velocity of the third fragment? Answer in units of m/s.
Answer:
Explanation:
Given
initial velocity of particle u=274 m/s
one Particle moves up with velocity of v=235 m/s
and other moves u=484 m/s towards east
let [tex]v_y[/tex] and [tex]v_x[/tex] be the velocity of third Particle is Y and x direction
conserving momentum in y direction
[tex]m(274)=\frac{m}{3}\times v_y+\frac{m}{3}\times 0+\frac{m}{3}\times 235[/tex]
[tex]v=587 m/s[/tex]
Now conserving momentum in x direction
[tex]m\times 0=\frac{m}{3}\times v_x+\frac{m}{3}\times 0+\frac{m}{3}\times 484[/tex]
[tex]v_x=-484 m/s[/tex]
Net Velocity of third Particle
[tex]v^2=v_x^2+v_y^2[/tex]
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{484^2+587^2}[/tex]
[tex]v=760.80 m/s[/tex]
To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. The magnitude of the velocity of the third fragment is 445 m/s.
Explanation:To find the magnitude of the velocity of the third fragment, we need to consider the conservation of momentum. According to the law of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. The momentum of an object is the product of its mass and velocity.
Let's assume the mass of each fragment is m. Since the first fragment continues to move upward with a speed of 235 m/s and the second fragment has a speed of 484 m/s and is moving east, we can write the momentum equation as:
m(235) + m(484) + m(v3) = m(274)
Simplifying the equation, we get:
719m + m(v3) = 274m
445m = - m(v3)
- 445m = m(v3)
Dividing both sides by m, we get:
-445 = v3
Therefore, the magnitude of the velocity of the third fragment is 445 m/s.
A 0.180 kg cube of ice (frozen water) is floating in glycerin. The glycerin is in a tall cylinder that has inside radius 5.50 cm. The level of the glycerin is well below the top of the cylinder.
A)If the ice completely melts, by what distance does the height of liquid in the cylinder change?
Express your answer with the appropriate units. Enter positive value if the surface of the water is above the original level of the glycerin before the ice melted and negative value if the surface of the water is below the original level of the glycerin.
Answer:
0.00389 m
Explanation:
Archimedes principle states that the buoyant force when a body is immersed in a liquid equals the weight of the liquid displaced
B = weight of the liquid displaced
density is defined as the mass divided by the volume of the substance and the units is in kg/m³
Density of glycerin = mass / volume
ρg × volume = mass of glycerin displaced
since the object was floating, the upthrust from the liquid equals the weight of the liquid
ρg × volume × g = mg
divide both side by g
ρg × volume = 0.180 where ρg (density of glycerin) = 1260 kg / m³
1260 × volume of glycerin displaced = 0.18 / 1260 = 0.000143 m³
volume of glycerin displaced = πr²h₁ where h₁ = of liquid displaced
πr²h₁ = 0.000143
h₁ = 0.000143 / ( 3.142 × 0.055² ) = 0.01505 m
when the ice completely melted, it will displaces liquid equal to it own volume
density of water = mass of water /volume
1000 = 0.18 / v
v = 0.18 / 1000 = 0.00018 m³
volume = πr²h₂ where h₂ = height of the melted water
πr²h₂ = 0.00018
h₂ = 0.00018 / ( 3.142 × 0.055²) = 0.01894 m
change in height of the liquid = h₂ - h₁ = 0.01894 m - 0.01505 m = 0.00389 m
When the 0.180 kg ice cube melts, it will form water that spreads out in the glycerin-filled cylinder, raising the level by approximately 1.89 cm.
Explanation:To solve this problem, we first need to understand that when the ice cube melts, it will not change the overall volume of liquid in the cylinder. In other words, the total volume before the ice melts (volume of glycerin + volume of ice) is equal to the total volume after the ice melts (volume of glycerin +volume of water).
The volume of the water, resulting from the melted ice, can be calculated using the mass of the ice cube and the density of water. The volume (V) is equal to the mass (m) divided by the density (ρ). For ice, m = 0.180 kg and ρ = 1000 kg/m³ (the density of water), which gives V = 0.00018 m³ or 180 cm³.
This water spreads out in the cylinder, raising the level. The height (h) to which it raises can be found by dividing the volume of the water (V) with the cross-sectional area (A) of the cylinder (h = V/A). The area of the cylinder is πr², where r is the radius, so A = π * (5.5 cm)² = 95.03 cm². Thus, h = 180 cm³ / 95.03 cm² = 1.89 cm. Consequently, when the ice cube melts, the glycerin's height will increase by approximately 1.89 cm.
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An alternating current is set up in an LRC circuit.
For which of the following circuit elements are the current and voltage in phase?
A) inductor only
B) resistor only
C) capacitor only
D) resistor and capacitor only
E) inductor, resistor, and capacitor
Answer:
(B) Resistor only
Explanation:
Alternating Current: These are currents that changes periodically with time.
An LRC Ac circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, connected in series.
In a purely resistive circuit, current and voltage are in phase.
In a purely capacitive circuit, the current leads the voltage by π/2
In a purely inductive circuit, the current lags the voltage by π/2.
Therefore when a alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.
The right option is (B) Resistor only.
The circuit elements are the current and voltage in phase in:
B) Resistor only
What is Alternating current?These are flows that changes occasionally with time. A LRC AC circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, associated in series. In an absolutely resistive circuit, current and voltage are in stage. In an absolutely capacitive circuit, the current leads the voltage by π/2. In an absolutely inductive circuit, the current slacks the voltage by π/2.
Thus, when an alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.
So, correct option is (B).
Find more information about Alternating current here:
brainly.com/question/10715323
What is the force of a 500kg car that is accelerating at 20m /sex2?
Answer:10000N
Explanation:formuler for calculating force is given by F=Ma
M(mass)=500kg
a(acceleration)=20m/s^2
Therefore by substitution we have F=500*20
F=10000N