Answer
given,
acceleration, a = 5 g = 5 x 9.8 = 49 m/s²
speed of sound, v = 331 m/s
speed , v = 3 Mach = 3 x 331 m/s = 993 m/s
time, t = 5.2 s
a) Calculating the time period of black out.
time required by the aircraft to black out
using equation of motion
v = u + a t
initial velocity = 0 m/s
993 = 0 + 49 t
t = 20.26 s
the required time to reach 3 Mach speed is more than the given time hence, pilot will black out.
b) now, calculating the maximum speed it can reach in 5.2 s
using equation of motion
v = u + a t
initial velocity = 0 m/s
v = 0 + 49 x 5.2
v = 254.8 m/s
The total magnification of a specimen being viewed with a 10X ocular lens and a 40X objective lens is _____.
Answer:
total magnification = 400 X
Explanation:
given data
ocular lens = 10 X
objective lens = 40 X
to find out
total magnification
solution
we know that total magnification is express as
total magnification = Objective magnification × ocular magnification .................1
put here value we get
total magnification = 10 × 40
total magnification = 400 X
The total magnification for a microscope with a 10X ocular lens and a 40X objective lens is 400X.
Explanation:When using a microscope, the total magnification can be found by multiplying the magnification of the ocular lens (also known as the eyepiece) by the magnification of the objective lens. Thus, in your case, the total magnification of the specimen being viewed with a 10X ocular lens and a 40X objective lens would be 400X. This is because 10 times 40 equals 400. Therefore, you are observing the specimen at 400 times its actual size.
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For every factor of 10 difference in Keq' (e.g., going from 102 to 10, or from 10–3 to 10–4), what is the difference in standard free energy change, ΔG°' (kJ/mol)?
Answer:
The standard free energy is the difference between that of products and reactants. The value of G has a greater influence on the reaction been considered.
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)if Delta -G = +ve ( the reaction is non-spontaneous)if Delta-G = 0 ( the reaction is at equilibrium)Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.
For each factor of 10 change in Keq, there is a corresponding change of approximately 5.7 kJ/mol in standard free energy change, ΔG°', at standard biochemical conditions (298.15 K).
Explanation:For every factor of 10 difference in the equilibrium constant, Keq, there is a corresponding change in the standard free energy change, ΔG°', as calculated using the relationship between ΔG°' and Keq:
ΔG°' = -RT ln Keq, where R is the universal gas constant and T is the temperature in Kelvin.
Given that R = 8.314 J/mol·K and at standard biochemical conditions T = 298.15 K, a tenfold change in Keq results in a change of (8.314 J/mol·K x 298.15 K x ln(10)), which equates to approximately 5.7 kJ/mol. Therefore, for example, increasing Keq from 102 to 103 would decrease ΔG°' by 5.7 kJ/mol, making the reaction more product-favored under standard conditions. Conversely, decreasing Keq from 10-3 to 10-4 would increase ΔG°' by 5.7 kJ/mol, making the reaction less product-favored under standard conditions.
The average lifetime of a pi meson in its own frame of reference (i.e., the proper lifetime) is 2.6 10-8 s. (a) If the meson moves with a speed of 0.88c, what is its mean lifetime as measured by an observer on Earth? (b) What is the average distance it travels before decaying, as measured by an observer on Earth? (c) What distance would it travel if time dilation did not occur?
Final answer:
a) The measured lifetime of the pi meson as observed by an Earth observer is longer than its proper lifetime due to time dilation. b) The average distance the meson travels before decaying can be calculated using the velocity and measured lifetime. c) If time dilation did not occur, the distance the meson would travel can be calculated using the velocity and proper lifetime.
Explanation:
a) According to time dilation, the average lifetime of the pi meson as measured by an observer on Earth is longer than its proper lifetime. To calculate it, we use the formula for time dilation: t' = t / γ, where t' is the measured lifetime, t is the proper lifetime, and γ is the Lorentz factor. Given that the proper lifetime is 2.6 × 10-8 s and the speed of the meson is 0.88c, we can calculate γ using the formula γ = 1 / √(1 - v2/c2). After calculating γ, we can substitute it into the time dilation formula to find the measured lifetime on Earth.
b) To calculate the average distance the meson travels before decaying, we use the formula d = v · t', where d is the distance, v is the velocity, and t' is the measured lifetime. We can substitute the values given in part a) to find the distance traveled by the meson.
c) If time dilation did not occur, the distance the meson would travel can be calculated using the formula d = v · t, where d is the distance, v is the velocity, and t is the proper lifetime. By substituting the values given in part a) into this formula, we can find the distance the meson would travel in the absence of time dilation.
How long (in s) will it take an 875 kg car with a useful power output of 42.0 hp (1 hp = 746 W) to reach a speed of 17.0 m/s, neglecting friction? (Assume the car starts from rest.)
Answer:
The time taken by the car with a useful power output is 4.03 seconds.
Explanation:
Given that,
Mass of the car, m = 875 kg
Initial speed of the car, u = 0
Final speed of the car, v = 17 m/s
Power of the car, P = 42 hp = 31332 W
The power output of the car is given by the work done per unit time. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
Initial kinetic energy of the car will be 0 as it is at rest.
Final kinetic energy of the car is :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]K=\dfrac{1}{2}\times 875\times (17)^2[/tex]
K = 126437.5
As per the work energy theorem, the change in kinetic energy of the object is equal to the work done.
So,
[tex]P=\dfrac{W}{t}[/tex]
[tex]t=\dfrac{W}{P}[/tex]
[tex]t=\dfrac{126437.5}{31332}[/tex]
t = 4.03 seconds
So, the time taken by the car with a useful power output is 4.03 seconds. Hence, this is the required solution.
Radiometric dating of a magnetic anomaly stripe of rock that is 225 km away from the mid-ocean ridge axis gives an age of 9 million years. Assuming a constant rate, seafloor spreading in this area occurs at a rate of __________.
A. 5 cm per year
B. 1,012.5 km per year
C. 20,000 cm per year
D. 50 km per year
The rate of seafloor spreading in this area, based on the given radiometric dating results and calculated assuming a constant rate, is 25 km/million years.
Explanation:
To calculate the rate of seafloor spreading, we divide the distance that the seafloor has spread by the amount of time it took. Given the radiometric dating results, we know that this distance (225 km) was covered in 9 million years.
So, the calculation would be: 225 km / 9 million years = 0.025 km/year. Or, to convert this figure to kilometers per million years, you'd multiply by 1 million, giving you a seafloor spreading rate of 25 km/million years.
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A truck is hoisted a certain distance in a garage and therefore has potential energy with respect to the floor. If it were lifted twice as high, it would have ____ as much potential energy.
Answer:
If the truck were lifted twice as high, it would have twice as much potential energy
Explanation:
Potential energy = mgh
where;
m is mass of the truck (kg)
g is acceleration due to gravity (m/s²)
h is the height above the floor.
Initial Potential Energy, E₁ = mgh₁
When the truck was lifted twice as high, New height (h₂) = 2h₁
New potential Energy, E₂ =mgh₂ = mg(2h₁)
E₂ = 2mgh₁
Recall, E₁ = mgh₁, then substitute in E₁ into E₂
E₂ = 2(mgh₁)
E₂ = 2E₁
Therefore, If the truck were lifted twice as high, it would have twice as much potential energy
The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are __________.
Answer:
The ability of water molecules to form hydrogen bonds with other water molecules and water's ability to dissolve substances that have charges or partial charges are due to water's partial charges.
Explanation:
The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.
Two canoes are touching and at rest on a lake. The occupants push away from each other in opposite directions, giving canoe 1 a speed of 0.56 m/s and canoe 2 a speed of 0.45 m/s .
If the mass of canoe 1 is 320 kg , what is the mass of canoe 2?
Answer:
398.22 kg
Explanation:
[tex]m_1[/tex] = Mass of canoe 1 = 320 kg
[tex]m_2[/tex] = Mass of canoe 2
[tex]v_1[/tex] = Velocity of canoe 1 = 0.56 m/s
[tex]v_2[/tex] = Velocity of canoe 2 = 0.45 m/s
In this system the linear momentum is conserved
[tex]m_1v_1=m_2v_2\\\Rightarrow m_2=\dfrac{m_1v_1}{v_2}\\\Rightarrow m_2=\dfrac{320\times 0.56}{0.45}\\\Rightarrow m_2=398.22\ kg[/tex]
The mass of the second canoe is 398.22 kg
Answer:
398.22 kg
Explanation:
mass of I canoe, m1 = 320 kg
initial velocity of both the canoe = 0
final velocity of I canoe, v1 = 0.56 m/s
final velocity of II canoe, v2 = - 0.45 m/s
Let the mass of second canoe is m2.
Use conservation of momentum
momentum before push = momentum after push
0 + 0 = m1 x v1 + m2 x v2
0 = 320 x 0.56 - m2 x 0.45
m2 = 398.22 kg
Thus, the mass of second canoe is 398.22 kg.
A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r(t) is to increase throughout the ball's flight?
The question involves the application of Newton's law to a ball's position over time, launched from a cannon at an angle. The x and y-axis positions can be calculated using horizontal and vertical principles of motion respectively. For the ball's distance (r) to increase throughout the flight, the launch angle needs to be 90°.
Explanation:The subject of this question relates to the application of Newton's second law in two dimensions, specifically in cases of projectile motion. The position of the ball fired from a cannon is determined by the initial velocity of the ball, the launch angle θ, and the acceleration due to gravity.
The x-component of the ball's position can be calculated using horizontal motion principles. Here, the x position x(t) = V₀cos(θ)t, where V₀ is the initial speed of the ball, cos(θ) is the cosine of the launch angle and t is the time.The y-component of the ball's position can be calculated using vertical motion principles. In this case, y(t) = V₀sin(θ)t - 0.5gt², where sin(θ) is the sine of the launch angle, g is the acceleration due to gravity, and t is the time. For r(t) to increase throughout the ball's flight, the launch angle θ needs to be 90°. This is because the maximum range or distance is attained when objects are projected at an angle of 45°. But the ball's range will start to decrease after this angle, while the height (r(t)) will continue to increase until 90°.Learn more about Projectile Motion here:
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At time t in seconds a particles distance s(t) in micrometers from a point is given by s(t)=e^t+2-sin(t) what is the average velocity of the particle from t=1 to t=5
Answer:
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
Explanation:
Given:
distance as a function of time is:
[tex]s=e^t+2-sin\ t[/tex]
Average velocity of the particle from time t=1s to t=5s can be given by the total distance covered divided by the total time consumed.
Position at t=1s:
[tex]s_1=e^1+2-sin\ 1[/tex]
[tex]s_1=3.8768\ m[/tex]
Position at t=5s:
[tex]s_5=e^5+2-sin\ 5[/tex]
[tex]s_5=151.3721\ m[/tex]
Therefore the distance covered during this time is:
[tex]\Delta s=s_5-s_1[/tex]
[tex]\Delta s=151.3721-3.8768[/tex]
[tex]\Delta s =147.4952\ m[/tex]
Now the average speed for the duration:
[tex]v_{avg}=\frac{\Delta s}{\Delta t}[/tex]
[tex]v_{avg}=\frac{147.4952}{5}[/tex]
[tex]v_{avg}=29.499\ m.s^{-1}[/tex]
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. What impulse was given to the ball by the floor?
Answer:
1.0752 kgm/s
Explanation:
Considering when the drop was dropped from rest from a height,
mass of the ball, m = 0.120 kg
height, h = - 1.25 m
the initial velocity, u = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)[/tex]
[tex]V^{2} = 24.5 m/s[/tex]
[tex]V = \sqrt{24.5} \ m/s[/tex]
[tex]V = 4.95 \ m/s[/tex]
V = ± 4.95 m/s
V = - 4.95 m/s
Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s
Considering when the ball rebounds from the floor,
assume the mass of the ball still remain, m = 0.120 kg
height, h = 0.820 m
the final velocity, v = 0 m/s
the acceleration due to gravity, g = - 9.8 m/s²
From equation of motion
[tex]V^{2} = U^{2} + 2gh[/tex]
Substituting the values,
[tex]0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)[/tex]
[tex]0 = U^{2} - 16.072 m/s[/tex]
[tex]U^{2} = 16.072 m/s[/tex]
[tex]U = \sqrt{16.072} \ m/s[/tex]
U = ± 4.01 m/s
U = + 4.01 m/s
Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s
From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.
[tex]F = \frac{mv - mu}{t}[/tex]
[tex]F.t = m(v - u)[/tex]
⇒ Impulse = Change in momentum
To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,
∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )
F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )
F.t = 0.120 kg × 8.96 m/s
Impulse = 1.0752 kgm/s
The impulse given to the ball by the floor is 1.0752 kgm/s
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?
Answer:
The frequency is 302.05 Hz.
Explanation:
Given that,
Speed = 18.0 m/s
Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .
We need to calculate the frequency
Using formula of frequency
[tex]f'=f(\dfrac{v+v_{p}}{v-v_{s}})[/tex]
Where, f = frequency
v = speed of sound
[tex]v_{p}[/tex] = speed of passenger
[tex]v_{s}[/tex] = speed of source
Put the value into the formula
[tex]f'=262\times(\dfrac{344+18}{344-30})[/tex]
[tex]f'=302.05\ Hz[/tex]
Hence, The frequency is 302.05 Hz.
If a ball end mill cutter's stepover is kept at a fixed value but the diameter of the tool is increased, the scallop height increases as a consequence.Select one:a. Trueb. False
Answer: FALSE
Explanation:Ball End Mills are hemispherical tip used to cut and and shape/ milling rounded objects, such as the metal bearing grooves, slotting and pocketing.
Ball End Mill also called Ball nose end mills, the scallop height will not increase as a result of the increase in diameter. The diameter is the distance between one end of the round tool to another when taken from the middle.
A bakery measured the mass of a whole cake as 0.870 kg. A customer bought one slice of the cake and measured the mass of the slice as 0.1151 kg. What is the mass of the remaining cake, without the one slice?
Answer:
0.7549kg
Explanation:
The mass of the slice + mass of the remaining cake = total mass of cake.
mass of remaining cake = total mass of cake - the mass of the slice
total mass=0.870kg
mass of slice = 0.1151kg
mass of remaining cake = 0.870 - 0.1151
mass of remaining cake=0.7549kg
A 106 kg clock initially at rest on a horizontal floor requires a 670 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 557 N keeps it moving with a constant velocity.
a) what is the μs between the clock and the floor?.70
b) what is the μk between the clock and the floor?
Answer:
a) [tex]\mu_s=0.65[/tex]
b) [tex]\mu_k=0.54[/tex]
Explanation:
Static friction is when the body is at rest or is about to move while kinetic friction is when the body is already in motion. According to Newton's second law:
[tex]\sum F_y:N=mg\\\sum F_x:F_f=F_x[/tex]
a) In this case, the static friction must be equal to the horizontal force to set the clock in motion:
[tex]F_f=\mu_sN=\mu_smg\\\mu_smg=F_x\\\mu_s=\frac{F_x}{mg}\\\mu_s=\frac{670}{106kg(9.8\frac{m}{s^2})}\\\mu_s=0.65[/tex]
b) In this case, the kinetic friction is equal to the horizontal force that keep the clock moving with constant velocity:
[tex]\mu_kmg=F_x'\\\mu_k=\frac{F_x'}{mg}\\\mu_k=\frac{557}{106kg(9.8\frac{m}{s^2})}\\\mu_k=0.54[/tex]
What is the magnitude of the net force ona(n) 97 kg driver operating a dragster as it accelerates horizontally along a straight line from rest to 40 m/s in 5 s?
Answer:
Force acting on the driver will be 776 N
Explanation:
We have given mass of the driver m = 97 kg
It starts from rest so initial velocity u = 0 m /sec
And reaches to a velocity of 40 m /sec in 5 sec
So final velocity v = 40 m /sec
And time taken t = 5 sec
From first equation of motion v = u +at
So [tex]40=0+a\times 5[/tex]
[tex]a=8m/sec^2[/tex]
Now we have to find the force acting the driver
From newtons law we know that F = ma
So force F = 97×8 = 776 N
So force acting on the driver will be 776 N
Use a(t) = -32 feet per second per second as the acceleration due to gravity. (Neglect air resistance.) -With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately 550 feet)?
Answer:
u= 187.61 ft/s
Explanation:
Given that
g= - 32 ft/s²
The maximum height ,h= 550 ft
Lets take the initial velocity = u ft/s
We know that
v²=u² + 2 g s
v=final speed ,u=initial speed ,s=height
When the object reach at the maximum height then the final speed of the object will become zero.
That is why
u²= 2 x 32 x 550
u²= 35200
u= 187.61 ft/s
That is why the initial speed will be 187.61 ft/s
Answer:
Explanation:
a = - 32 ft/s²
h = 550 ft
Let the initial velocity is u. The velocity at maximum height is zero.
use third equation of motion
v² = u² + 2 a h
0 = u² - 2 x 32 x 550
u² = 35200
u = 187.62 ft/s
A drawing of electric field lines will immediately reveal (1) the relative magnitude of different charges (proportional to the number of lines that begin or end on each), (2) the sign of different charges (since lines go into negative charges and come out of positive charges), (3) the relative magnitude of the electric field at any point (because the magnitude of the field is proportional to how closely spaced the field lines are), and (4) the symmetry of the charge distribution (since they match the symmetry of the underlying field).
Complete Question:
The complete question is on the all the uploaded image
Answer:
sfjjA)rule 1,3,5 are brokenB)rule 2,3 are brokenC) rule 2 are brokenD) rule 2,3 are brokenE)rule 2 are brokenF)noneG) rule 3,4 are brokenH) rule 2,4,5 are brokenExplanation:
This solution were gotten by examining the diagrams and figuring out the rules that are broken
For A
we see that the two charges are positive so rule 1 is broken,on the same diagram we see that there is no radial symmetry close to the charge so rule 3 is broken then looking again at the diagram A we can see that the electric field is not tangent to any electric line at any point hence rule 5 is broken
For B:
We see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For C:
We see that the number of line is not proportional to the magnitude of the charge hence rule 2 is broken.
For D:
we see that the number of lines are not proportional to the magnitude of the charge hence rule 2 has been broken
secondly looking again at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
For E:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
For F
Looking at the diagram we see that none of the rules are broken
For G:
looking at the diagram we see that the line are not uniformly distributed near the charge hence rule 3 is broken
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
For H:
We see that the number of line is not proportional to he magnitude of the charge hence rule 2 is broken.
taking another look at the diagram we see that the spacing of the lines are not indicating the magnitude of the charge hence rule 4 is broken
Looking again at the diagram we see that the at any point on the electric field line that the electric field itself is not tangent to the electric field line hence rule 5 is broken
Radio Station: What is the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz? Remember that λ = c/f, where λ is the wavelength, f is the frequency and c is the speed of light (c = 300000 km/s)
Answer: The wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
Explanation:
To calculate the wavelength of light, we use the equation:
[tex]\lambda=\frac{c}{\nu}[/tex]
where,
= wavelength of the light = ?
c = speed of light = 300000 km/s = [tex]3\times 10^8m/s[/tex] (1km=1000m)
[tex]\nu[/tex] = frequency of light = 100kHz = 100000Hz or (1kHz=1000Hz)
[tex]\lambda=\frac{3\times 10^8m/s}{100000s^{-1}}[/tex]
[tex]\lambda=3000m[/tex]
Thus the wavelength of a radio station photon from an AM radio station that broadcast at 1000 kilo-hertz is 3000 m.
A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft long, and its modulus of elasticity is 30,000,000 psi. if the strain in the diagonal member is measured as 0.001200 in./in., determine:
Answer:
axial stress in the diagonal bar =36,000 psi
Explanation:
Assuming we have to find axial stress
Given:
width of steel bar: 1.25 in.
height of the steel bar: 3 in
Length of the diagonal member = 20ft
modulus of elasticity E= 30,000,000 psi
strain in the diagonal member ε = 0.001200 in/in
Therefore, axial stress in the diagonal bar σ = E×ε
= 30,000,000 psi× 0.001200 in/in =36,000 psi
The stress in the bridge truss diagonal tension member is computed to be 36000 psi, and the elongation of the member under this stress is calculated to be approximately 0.384 inches.
Explanation:The subject of the question is in the field of Engineering, particularly dealing with the stress and strain effects in a steel structural member of a bridge.
The stress in the member can be calculated using stress and strain relationship along with Young's modulus, given by the formula Stress = Strain x Young's Modulus. Given that strain is 0.001200 in./in. and Young's Modulus (E) is 30,000,000 psi, stress (σ) in the member is σ = E x Strain = 30,000,000 psi x 0.001200 in./in. = 36000 psi.
The steel bar's cross-sectional area can also be calculated as Area = Width x Depth = 1.25 in. x 3 in. = 3.75 in^2. The change in length (∆L) of the member under this stress can be calculated using the formula ∆L = (Stress x Original length) / (Young's Modulus x Area) = (36000 psi x 20 ft) / (30,000,000 psi x 3.75 in^2) = 0.032 ft or 0.384 in.
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____ is the process of sending data over a signal by varying either its amplitude, frequency, or phase.
Answer:
Modulation
Explanation:
Modulation is the the process whereby a waveform has one or more of it's properties varied. Amplitude, frequency and phase are all properties of a waveform which can be varied hence Modulation is the process of sending data over a signal by varying either its amplitude, frequency or phase.
Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., what is the potential drop across the conductor?
Answer:
3.53 V
Explanation:
Electric charge: The is the rate of flow of electric charge along a conductor.
The S.I unit of electric charge is C.
Mathematically it is expressed as,
Q = It ............................ Equation 1
Where Q = electric charge, I = current, t = time.
I = Q/t.......................... Equation 2
From the question, charge flows through the conductor at the rate of 420 C/mim
Which means in 1 min, 420 C of charge flows through the conductor.
Hence,
Q = 420 C, t = 1 min = 60 seconds
Substitute into equation 2
I = 420/60
I =7 A
Also
P = VI......................... Equation 3
Where P = power, V = potential drop, I = current.
V = P/I................... Equation 4
Note: Power = Energy/time
From the question, P = 742/30 = 24.733 W. and I = 7 A.
Substitute these values into equation 4
V = 24.733/7
V = 3.53 V
Hence the potential drop across the conductor = 3.53 V
The potential drop across the conductor is 3.54 V
The rate of flow of electric charge along a conductor per unit time is known as current.
Mathematically it is expressed as,
As we know that
Power, [tex]P = VI[/tex]
Where P is power, V is potential drop and I is current.
Also,
It is given that, energy = 742 J and time = 30s
Power [tex]=\frac{742}{30} =24.74watt[/tex]
Substituting the value of power in equation [tex]P=VI[/tex]
[tex]V=\frac{P}{I} =\frac{24.74}{7}=3.54V[/tex]
Thus, the potential drop across the conductor is 3.54 V
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During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the slope, and 1.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.52 km x 0.52 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 5.5 m2 area of the valley floor?
Answer:
93328 kg
Explanation:
[tex]\rho[/tex] = Density of mud = 1900 kg/m³
Volume of the cuboid is given by
[tex]V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3[/tex]
Volume is also given by
[tex]V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m[/tex]
The length of the cuboid is 0.89312 m
For the small volume
[tex]V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3[/tex]
Mass is given by
[tex]m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg[/tex]
The mass of the mud is 93328 kg
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its mass is 102 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
Explanation:
Power is the ratio of energy to time.
Here we need to consider kinetic energy,
Mass, m = 102 kg
Initial velocity = 0 m/s
Final velocity = 16.2 m/s
Time, t = 2.4 s
Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J
Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J
Change in energy = Final kinetic energy - Initial kinetic energy
Change in energy = 13384.44 - 0
Change in energy = 13384.44 J
Power = 13384.44 ÷ 2.4 = 5576.85 W = 7.48 hp
Power of cheetah is 5576.85 W = 7.48 hp
A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping force Fx= −bvx acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s. Calculate the magnitude of the damping constant b. Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures.
Answer:
0.02896 kg/s
Explanation:
[tex]A_1[/tex] = Initial displacement = 0.5 m
[tex]A21[/tex] = Final displacement = 0.1 m
t = Time taken = 0.5 s
m = Mass of object = 45 g
Displacement is given by
[tex]x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)[/tex]
At maximum displacement
[tex]cos(\omega t+\phi)=1[/tex]
[tex]\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s[/tex]
The magnitude of the damping coefficient is 0.02896 kg/s
Final answer:
To find the damping constant b for a damped harmonic motion scenario with a given change in amplitude over time, you employ the decay of amplitude formula in damped harmonic motion, then rearrange and solve for b.
Explanation:
To calculate the magnitude of the damping constant b for a 45.0 g hard-boiled egg moving at the end of a spring with a force constant of 25.0 N/m given a decrease in amplitude from 0.500 m to 0.100 m over 5.00 seconds, we apply the principle of damped harmonic motion. The damping force is given by Fx= −bvx, where b is the damping coefficient we wish to find. From the information provided, it's understood that this is a case of exponentially decaying amplitude in harmonic motion.
The equation that relates the exponential decay of amplitude in damped harmonic motion is A(t) = A0e−(bt/2m), where A0 is the initial amplitude, A(t) is the amplitude at time t, and m is the mass of the object. By re-arranging this equation to solve for b, and substituting the given values A0=0.500 m, A(t)=0.100 m, t=5.00 s, and m=0.045 kg (since 45.0 g=0.045 kg), we can calculate b.
The rearranged equation for b would be b = 2m[ln(A0/A(t))]/t. Substituting the values gives b = 2(0.045)[ln(0.500/0.100)]/5.00, which after calculation yields the magnitude of b.
A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the springconstant?
Answer:
a)F=698.83 N
b)K=8221.56 N/m
Explanation:
Given that
mass ,m = 0.12 kg
Amplitude ,A= 8.5 cm
time period ,T = 0.2 s
We know that
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]{\omega}=\dfrac{2\pi }{0.2}\ rad/s[/tex]
[tex]{\omega}=31.41\ rad/s[/tex]
We know that
[tex]{\omega}^2=m\ K[/tex]
K=Spring constant
[tex]K=\dfrac{\omega^2}{m}[/tex]
[tex]K=\dfrac{31.41^2}{0.12}\ N/m[/tex]
K=8221.56 N/m
The maximum force F
F= K A
F= 8221.56 x 0.085 N
F=698.83 N
a)F=698.83 N
b)K=8221.56 N/m
Final answer:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we use the equation F = -kx. To find the spring constant, we can rearrange the equation F = -kx.
Explanation:
To find the magnitude of the maximum force acting on the body in simple harmonic motion, we can use the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. In this case, we are given the amplitude A = 8.5 cm and the mass m = 0.12 kg. The displacement x at the maximum amplitude is equal to the amplitude, so x = A.
Plugging in the values, we get F = -(k)(A).
(a) To find the magnitude of the maximum force, we need to find the spring constant k. We can use the equation T = 2π√(m/k), where T is the period. Plugging in the values, T = 0.20 s and m = 0.12 kg, we can solve for k.
(b) To find the spring constant, we can rearrange the equation F = -kx to solve for k. Plugging in the values, F = 2.00 mg and x = A, we can solve for k.
A Sense of Proportion: The Sun is roughly 100 times the diameter of Earth. If you built a model solar system with a golf ball (diameter of 1.68 inches) for Earth, how big would the model Sun be?
Explanation:
Diameter of Sun = 100 x Diameter of Earth.
We are creating a model where Earth is same as golf ball,
Diameter of Earth in model = Diameter of golf ball
Diameter of Earth in model = 1.68 inches
Diameter of Sun in model = 100 x Diameter of Earth in model
Diameter of Sun in model = 100 x 1.68
Diameter of Sun in model = 168 inches = 14 feet
Diameter of Sun in model = 168 inches = 14 feet = 4.27 m
Friction pulls directly against the direction of motion (at 180) of a sled, and does -55.2 J of work while the sled moves 8.98 m. What is the magnitude (+) of the friction force?(unit=N)
The magnitude of the force of friction is 6.1 N
Explanation:
The work done by a force when moving an object is given by the equation:
[tex]W=Fd cos \theta[/tex]
where :
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem, we have:
W = -55.2 J (work done by the force of friction)
d = 8.98 m (displacement of the sled)
[tex]\theta=180^{\circ}[/tex] (because the force of friction acts opposite to the direction of motion)
By solving the equation for F, we find the magnitude of the force of friction on the sled:
[tex]F=\frac{W}{d cos \theta}=\frac{-55.2}{(8.98)(cos 180^{\circ})}=6.1 N[/tex]
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Answer:
6.1
Explanation:
Acellus
When two charge producers with different surface materials are rubbed together to create a charge imbalance. True or False
Answer:
True
Explanation:
Answer:
The answer is TRUE
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As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it consumed). However, the pounce itself only required 4 units of kinetic energy. How many units of energy were dissipated as heat?
The cat used 10 units of potential energy to pounce but only 4 units were converted meaningfully into kinetic energy of the pounce. The remaining 6 units were dissipated as heat.
Explanation:When the cat pounces on a mouse, the energy needed for this action is less than the total potential energy consumed. The cat's body uses 10 units of potential energy, but the pounce only required 4 units of kinetic energy. The difference between these two values signifies the amount of energy dissipated as heat. Therefore, we can perform the subtraction: 10 units (total potential energy) - 4 units (kinetic energy used) = 6 units. Thus, "6 units of energy were dissipated as heat."
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