A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimeter containing 75 g of water at 20°C.
The calorimeter is constructed of a material that has a specific heat of 0.10 cal/ g⋅°C.
When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.
a. 114°C
b. 72°C
c. 64°C
d. 37°C

Final answer:

The work done on the refrigerator during its trip down the ramp is 1350 J.

Explanation:

To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.

The displacement is given by the length of the ramp, which is 5.0 m.

The force applied is given as 270 N.

To calculate the work done, we use the formula:

Work = Force x Displacement x Cosine(angle)

In this case, the angle is 0 degrees since the force is applied horizontally.

Therefore, the work done on the refrigerator is:

Work = 270 N x 5.0 m x Cos(0°) = 1350 J.

Answer:

(D) 10 seconds

Explanation:

The Houston Methodist Hospital automated the emergency power supply system (EPSS) testing processes in order to ensure the safety of patients during an outage

There is an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds.

Answer:

D. 10 seconds

Explanation:

The Houston Methodist Hospital has eleven diesen powered generators that turn on at most 10 seconds after the normal power shuts down.

Considering that it is a hospital, this system is really important.

So the correct answer is:

D. 10 seconds

Answer:

1002.2688 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

h = The length of a string = 2 m

m = Mass of block = 1.6 kg

[tex]m_2[/tex] = Mass of bullet = 0.01 kg

Here, the potential energy of the fall will balance the kinetic energy of the bullet

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]

Velocity of block is 6.26418 m/s

As the momentum of system is conserved we have

[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]

The magnitude of velocity just before hitting the block is 1002.2688 m/s

Using Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target are respectively;

v1 = (m1•u1 - m2•u1)/(m1 + m2)

v1 = (m1•u1 - m2•u1)/(m1 + m2)v2 = (m1•u1 + m1•u2)/(m1 + m2)

Let us denote the following;

Mass of projectile; m1

Mass of target; m2

Initial velocity of projectile; u1

Final velocity of projectile; v1

Initial velocity of target; u2 = 0 m/s

Final velocity of target; v2

From conservation of momentum, we have;

(m1•u1) + (m2•u2) = (m1•v1) + (m2•v2)

Plugging in the relevant values;

(m1•u1) + (m2•0) = (m1•v1) + (m2•v2)

(m1•u1) = (m1•v1) + (m2•v2) - - - (eq 1)

From energy principle, we have;

u1 + v1 = u2 + v2

Thus;

Since u2 is zero, then we have;

u1 + v1 = v2 - - - (eq 2)

Put (u1 + v1) for v2 in eq 1 to get;

(m1•u1) = (m1•v1) + (m2(u1 + v1))

(m1•u1) = (m1•v1) + (m2•u1 + m2•v1) - - - (eq 3)

Let us make v1 the subject;

m1•u1 - m2•u1 = m1•v1 + m2•v1

v1(m1 + m2) = m1•u1 - m2•u1

v1 = (m1•u1 - m2•u1)/(m1 + m2)

Let us make v1 the subject in eq 2;

v1 = v2 - u2

Put v2 - u2 for v1 in eq 1 to get;

(m1•u1) = (m1(v2 - u2) + (m2•v2)

(m1•u1) = m1•v2 - m1•u2 + m2•v2

m1•u1 + m1•u2 = v2(m1 + m2)

v2 = (m1•u1 + m1•u2)/(m1 + m2)

Read more at; https://brainly.com/question/13994268

The final velocities of the projectile and the target after an elastic collision can be arrived at by solving the equations for Conservation of Momentum and Conservation of Kinetic Energy simultaneously. Here, the target's initial velocity is zero (it's at rest). By substituting this into the equations and solving them, we can find the final velocities.

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For this problem, we can solve the final velocities using the conservation laws.

Conservation of Momentum gives us:

m1*v1 = m1*v1' + m2*v2'     -------- (1)

V1' and V2' are the final velocities of the projectile and target, respectively.

Conservation of Kinetic Energy gives us:

1/2*m1*v1^2 = 1/2*m1*(v1')^2 + 1/2*m2*(v2')^2     -------- (2)

While solving these two equations, one should remember that m2 was initially at rest, so the momentum in the Y-direction is zero. The above-mentioned equations can be solved by substitution method or any other algebraic method to get the final velocities v1' and v2'.

https://brainly.com/question/33268757

#SPJ11

The problem is solved by applying the Ideal Gas Law and Dalton's Law of Partial Pressures. First, we calculate the mass of acetone condensed using these laws and then determine the rate at which acetone is vaporized using the given volumetric flow rate.

The given problem involves a number of gas law principles, but its main focus is on the application of the Ideal Gas Law and Dalton's Law of partial pressures. Initially, we calculate the moles of acetone in the feed using the Ideal Gas Law, and then we find out the moles of acetone in the effluent using Dalton's law. Subtracting gives us the moles of acetone condensed.

(a) Using Ideal Gas Law, we have PV=nRT. Hence, n (acetone, feed) = P (acetone, feed) * V(feed) /RT(feed). To find the moles of acetone in the effluent, we use Dalton's law and the Ideal Gas Law to get n (acetone, effluent) = P(acetone, effluent) * V (effluent) / RT (effluent). Subtracting moles in effluent from moles in feed gives moles condensed. Multiplying by the molar mass of acetone, we get mass of acetone condensed.

(b) The question tells us the volumetric flow rate of the gas leaving the condenser. Therefore, number of moles of acetone vaporized per hour can be calculated using the Ideal Gas Law and then can be converted into mass by using the molar mass of acetone. Hence, rate at which acetone is vaporized = moles (acetone, vaporized per hour) * molar mass (acetone).

https://brainly.com/question/34327119

#SPJ11

The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.

Mathematically the volume of this body is given as

[tex]V = lWh[/tex]

Where,

L = Length

W = Width

H = High

[tex]V = (12)(0.65)(13*10^{-2})[/tex]

[tex]V = 1.014m^3[/tex]

Note: The value given for the height was in centimeters, so it was transformed to meters.

Answer:

Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁

Explanation:

Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r)  exerted in x direction and reversal normal force N wall exerted on person.

                     [tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)

In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.

                     [tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)

from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.

Answer:negative Z direction

Explanation:

It is given that EM wave is travelling in negative y direction i.e. -[tex]\hat{j}[/tex]

Magnetic Field is in positive x direction i.e. [tex]\hat{i}[/tex]

We know that Electric field and magnetic field are perpendicular to each other and they are mutually perpendicular to direction of wave propagation.

Mathematically if Electric field is in negative z direction it will yield the direction of wave propagation

[tex]=(-\hat{k})\times \hat{i}[/tex]

[tex]=-\hat{j}[/tex]                            

Answer:

The right option is (d) substance undergoing a change of state

Explanation:

Latent Heat: Latent heat is the heat required to change the state of a substance without change in temperature. Latent heat is also known as hidden heat because the heat is not visible. The unit is Joules (J).

Latent heat is divided into two:

⇒ Latent Heat of fusion

⇒ Latent Heat of vaporization.

Latent Heat of fusion: This is the heat  energy required to convert a substance from its solid form to its liquid form without change in temperature. E.g (Ice) When ice is heated, its temperature rise steadily until a certain temperature is reached when the solid begins to melts.

Latent Heat of vaporization: This is the heat required to change a liquid substance to vapor without a change in temperature. The latent heat depend on the mass of the liquid and the nature of the liquid. E.g When water is heated from a known temperature  its boiling point (100°C) When more heat is supplied to its boiling temperature, it continue to boil without a change in temperature.

From The above, Latent heat brings about  a change of state of a substance at a steady temperature.

The right option is (d) substance undergoing a change of state

The substance that best represents a temperature remaining constant despite an inward heat flow is one that is undergoing a change of state, such as during the melting of ice or boiling of water. So, the correct option is D.

The best description of a substance in which the temperature remains constant while it is experiencing an inward heat flow is a substance undergoing a change of state (option d). This occurs during a process called phase transition.

Here's an example: when ice melts into water, it absorbs heat from the surroundings but the temperature stays constant at 0 degrees Celsius until all of the ice has melted.

Similarly, when water is boiling, its temperature holds steady at 100 degrees Celsius until all the water has transitioned into steam, despite the continued input of heat.

Learn more about Change of State here:

https://brainly.com/question/2273588

#SPJ3

Answer

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool

mass of the two person = 150 Kg

    weight of water level displaced is equal to the weight of person

    ρ V g = 2 m g

     [tex]V = \dfrac{2 m}{\rho}[/tex]

     [tex]V = \dfrac{2\times 150}{1000}[/tex]

            V = 0.3 m³

Area of pool = [tex]\dfrac{\pi}{4}d^2[/tex]

                    =[tex]\dfrac{\pi}{4}\times 6^2[/tex]

                    = 28.27 m²

height of the water rise

        [tex]h = \dfrac{V}{A}[/tex]

        [tex]h = \dfrac{0.3}{28.27}[/tex]

               h = 0.0106 m

pressure increased

           P = ρ g h

           P = 1000 x 10 x 0.0106

           P = 106.103 Pa

a. the absolute pressure at the bottom of the pool is approximately 116.02 kPa.

b. the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.

To find the absolute pressure at the bottom of the pool, we can use the hydrostatic pressure formula:

(a) Absolute pressure at a depth in a fluid:

P=P_0 +ρ⋅g⋅h

Where:

P is the absolute pressure at the given depth.

P_0 is the atmospheric pressure (which we'll assume to be 1 atm, or approximately 101.3 kPa).

ρ is the density of the fluid (water), which is about 1000 kg/m³.

g is the acceleration due to gravity, approximately 9.81 m/s².

h is the depth below the surface.

Given that the diameter of the pool is 6.00 m, the radius (r) is 3.00 m, and the depth (h) is 1.50 m, we can find the pressure at the bottom of the pool:

P=101.3kPa+(1000kg/m³)⋅(9.81m/s²)⋅(1.50m)

Solve for P:

P≈101.3kPa+14,715Pa≈116,015Pa≈116.02kPa

So, the absolute pressure at the bottom of the pool is approximately 116.02 kPa.

(b) To find the pressure increase at the bottom of the pool after two persons with a combined mass of 150 kg enter and float, we need to consider the additional weight of the water displaced by the persons. This additional weight will create an increase in pressure at the bottom of the pool.

The buoyant force (Fb) on an object submerged in a fluid is given by Archimedes' principle:

Fb =ρ⋅V⋅g

Where:

Fb is the buoyant force.

ρ is the density of the fluid (water), which is 1000 kg/m³.

V is the volume of water displaced, which is equal to the volume of the submerged part of the persons.

g is the acceleration due to gravity (9.81 m/s²).

The volume of water displaced is equal to the volume of the persons submerged, which is given by the cross-sectional area (A) of the pool base times the depth (h) to which they are submerged:

V=A⋅h

A=π⋅r^2

Given that the radius (r) is 3.00 m and the depth (h) is 1.50 m, we can calculate the area and then the volume:

A=π⋅(3.00m)^2 ≈28.27m²

V=28.27m²⋅1.50m≈42.41m³

Now, we can calculate the buoyant force:

Fb=(1000kg/m³)⋅(42.41m³)⋅(9.81m/s²)≈416,068N

Since pressure is force per unit area, we need to calculate the additional pressure (ΔP) at the bottom of the pool:

ΔP= Fb/A

​A=π⋅(3.00m)^2 ≈28.27m²

ΔP= 416,068N / 28.27m² ≈14,699Pa≈0.147kPa

So, the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.

Learn more about Pressure at the bottom of a pool here:

https://brainly.com/question/29650777

#SPJ12

Answer:

w = 7.89 10⁻² rad/s

Explanation:

We will solve this exercise with the conservation of the annular moment, let's write it in two moments

Initial. With the insect in the center

      L₀ = I w₀

End with the bug on the edge

     [tex]L_{f}[/tex]= I w +  [tex]I_{bug}[/tex] w

The moments of inertia are

For a rod

       I = 1/3 M L²

For the insect, taken as a particle

       I = m L²

The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved

      L₀ =  [tex]L_{f}[/tex]

      I w₀ = I w + [tex]I_{bug}[/tex] w

      w = I / (I +  [tex]I_{bug}[/tex]) w₀

      w = I / (I + m L²) w₀

Let's calculate

      w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)²   0.185

      w = 1.43 10⁻³ / 3.352 10³ 0.185

      w = 7.89 10⁻² rad/s

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

[tex]v_f[/tex] = Final velocity

[tex]v_i[/tex] = Initial velocity

[tex]t_f[/tex] = Final time

[tex]t_i[/tex] = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

[tex]\Delta P=F\Delta t[/tex]

Force

[tex]F=mg[/tex]

So,

[tex]m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s[/tex]

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

[tex]K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m[/tex]

The maximum height above your hand reached by the ball is 5.93505 m

The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.

The net force using the momentum principle can be given as,

[tex]F_{net}=\dfrac{\Delta P}{\Delta t}[/tex]

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,

[tex]F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}[/tex]

The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.

Therefore, the above equation can be written as,

[tex]mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}\\[/tex]

As the final velocity is zero and initial time is also zero. Therefore,

[tex](-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s[/tex]

Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,

[tex]\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m[/tex]

Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

Learn more about the momentum here;

https://brainly.com/question/7538238

To calculate the time it takes for the coffee pot to bring the water to a boil, we need to determine the amount of heat required and use the power of the coffee pot. By calculating the heat required for both the water and the aluminum pot, and using the power of the coffee pot, we can determine that it takes approximately 96.077 seconds for the coffee pot to bring the water to a boil.

To determine how long it takes for the coffee pot to bring the water to a boil, we need to calculate the amount of heat required. The heat required to raise the temperature of the water from 15?C to 100?C is given by the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Using the values given, we can calculate the heat required.

The heat required for the water is Q = (0.85 kg)(4186 J/kg?C)(100?C - 15?C) = 43707.5 J. The heat required for the aluminum coffee pot is Q = (0.36 kg)(900 J/kg?C)(100?C - 15?C) = 28350 J. Adding the two heats together gives us a total heat of 72057.5 J.

The power of the coffee pot is given as 750 W. Power is defined as the rate at which work is done or the rate at which energy is transferred. So, we can use the formula P = E/t, where P is power, E is energy, and t is time. Rearranging the formula, we can solve for t:

t = E/P = 72057.5 J / 750 W = 96.077 seconds.

https://brainly.com/question/30260910

#SPJ12

Answer:

a. T₂ = 901.43 ° K

b. T₂ = 843.85 ° K

Explanation:

Given

η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K

Using the equation that describe the experiment of heat

Q = η * Cv * ΔT

a.

Nitrogen   Cv =  20.76 J / mol ° K  

ΔT = T₂ - T₁

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K

T₂ = 901.43 ° K

b.

Nitrogen   Cp = 29.07 J / mol ° K

T₂ = [ Q / ( η * Cv) ] + T₁

T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K

T₂ = 843.85 ° K

The Final temperature of the air when the volume of the cylinder is constant is 700 K.

To find the final temperature of the air when the volume of the cylinder is constant, we can use the ideal gas law equation:

P₁V₁/T₁ = P₂V₂/T₂

Since the volume is constant, V₁ = V₂.

The initial pressure (P₁) and temperature (T₁) are given, and the final pressure (P₂) is also given (3.40×106 Pa).

We can rearrange the equation to solve for T₂:

T₂ = (P₂T₁) / P₁

Substituting the given values, we have:

T₂ = (3.40×106 Pa * 700 K) / (3.40×106 Pa)

T₂ = 700 K

So, the final temperature of the air is 700 K when the volume of the cylinder is constant.

Learn more about Final temperature here:

https://brainly.com/question/35133820

#SPJ6

Answer:

the question is incomplete, below is the complete question

"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.

a.What is the object's velocity, in meters per second, at time t = 2.9?

b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.  

c. What is the magnitude of the object's maximum acceleration, in meters per second squared?

d.What is the magnitude of the object's maximum velocity, in meters per second?"

a.[tex]v(t)==24.1m/s[/tex]

b.[tex]a(t)=3.79m/s^{2}[/tex]

c.[tex]a_{max}=106.48m/s^{2}[/tex]

d.[tex]v_{max}=24.2m/s[/tex]

Explanation:

the gneral expression for the displacement of object in simple harmonic motion is represented by

[tex]x(t)=Acos(wt- \alpha)\\[/tex]

while the velocity is express as

[tex]v(t)=-Awcos(4.4t-1.8)\\[/tex]

and the acceleration is

[tex]a(t)=-aw^{2}cos(wt- \alpha )\\[/tex]

Note: the angle is in radians

The expression for the displacement from the question is [tex]x(t)=5.5cos(4.4t-1.8)\\[/tex]

comparing, A=5.5, w=4.4,α=1.8

a.To determine the object velocity at t=2.9secs,

we substitute for t in the velocity equation

[tex]v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\[/tex]

[tex]v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s[/tex]

b.To determine the object acceleration at t=2.9secs,

we substitute for t in the acceleration equation

[tex]a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\[/tex]

[tex]a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}[/tex]

c. The acceleration is maximum when the displacement equals the amplitude. hence  magnitude of the object acceleration is

[tex]a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}[/tex]

d.The maximum velocity is expressed as

[tex]v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s[/tex]

Answer:

Part A:

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B:

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C:

θ′B = tan⁻¹(1.0699) = 46.94°

Part D:

v′B = 1.246 m/s

Explanation:

Given:  

mA = 0.117 kg

vA = vAx = 2.80 m/s  

mB = 0.135 kg

vB = 0 m/s

θ′A = 30.0°

v′A = 2.10 m/s

Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:

mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B

⇒ v′B*cosθ′B  = 0.8505  ⇒ v′B = 0.8505/cosθ′B

then

0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B

⇒ v′B*sinθ′B  = 0.91   ⇒ v′B = 0.91/sinθ′B

if we apply

0.8505/cosθ′B = 0.91/sinθ′B

⇒ tanθ′B = 0.91/0.8505 = 1.0699

⇒  θ′B = tan⁻¹(1.0699) = 46.94°

Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:

if   v′B = 0.91/sinθ′B

⇒ v′B = 0.91/sin 46.94°

⇒ v′B = 1.246 m/s

Conservation of momentum can be used to solve the equations for the angle and the speed of ball B after the collision, using the initial conditions, final conditions, and trigonometric identities. This involves the application of physics concepts, combined with the mathematics of trigonometry.

Part A: The correct equation expressing the conservation of momentum for the components in the x direction would be option B, mAvA=mAv′Acosθ′A+mBv′Bcosθ′B. This equation shows that the initial momentum of ball A (mAvA) equals the sum of the momentum of ball A and ball B after the collision in the x direction.

Part B: For the components in the y direction, the right answer is C, 0=mAv′Asinθ′A−mBv′Bsinθ′B. Since ball B was initially at rest and ball A was moving along the x-axis, there was no momentum in the y direction before the collision. Therefore, the total momentum in the y direction after the collision should also be 0.

Part C and D: To solve these equations for the angle and the speed of ball B after the collision, you need to use these equations in combination with the conservation of kinetic energy formula (1/2*m*v^2) and the trigonometric identities. Detailed solution steps require knowledge of the involved mathematics.

https://brainly.com/question/33316833

#SPJ3

Answer:

169.4 m/s

Explanation:

Given that the angle of projectile is θ_1 =450°

The speed of body at maximum height is U cosθ_1 = 150 m/s

The angle in second trail is θ_2 =37°  

From the given data U cosθ_1 = 150 m/s

U = 150 m/s / cosθ_1

=  150m/s / cos45°

=212.13 m/s

The velocity of the projectile at maximum height in second trail= Ucos(θ_2)

=212.13 m/s×cos37°

=169.4 m/s

The velocity of a projectile at its highest point when fired at an angle to the horizontal remains the same if the initial speed is unchanged, regardless of the angle. Hence, even when changing the angle from 45 to 37 degrees, the velocity at the highest point would still be 150 m/s.

The student's question deals with the velocity of a projectile at the highest point in its trajectory. When a projectile is fired upward at an angle, its velocity at the highest point of its trajectory is only composed of the horizontal component because the vertical component of the velocity becomes zero at that point.

In the given scenario, when the projectile is shot at 45 degrees to the horizontal, the speed at the highest point is given as 150 m/s. This speed represents the horizontal component since there's no vertical component at the highest point. When the angle is changed to 37 degrees, the horizontal component of the initial velocity is calculated using the cosine component of the initial speed, hence the velocity at the highest point remains the same as when it was fired at 45 degrees, provided that the initial speed is unchanged.

Therefore, in the second trial, with the angle at 37 degrees, the velocity of the projectile at its highest point would still be 150 m/s, because the horizontal component of the velocity is not affected by the change in angle.

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

[tex]\tau = Fr[/tex]

Where,

F = Force

r = Radius

Replacing we have that,

[tex]\tau = Fr[/tex]

[tex]\tau = 21cm (\frac{1m}{100cm})* 550N[/tex]

[tex]\tau = 11.55Nm[/tex]

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

[tex]I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2[/tex]

[tex]I = 0.394kg\cdot m^2[/tex]

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

[tex]\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}[/tex]

[tex]\alpha = \frac{11.55}{0.394}[/tex]

[tex]\alpha = 29.3 rad/s^2[/tex]

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians [tex](\pi / 3)[/tex], therefore

[tex]W = \tau \theta[/tex]

[tex]W = 11.5* \frac{\pi}{3}[/tex]

[tex]W = 12.09J[/tex]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE      (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² =  m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s²  = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]

is the enthalpy change

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

[tex]\Delta T=100-21\ ^0C=79\ ^0C[/tex]

So,  

[tex]\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ[/tex]

Heat Supplied [tex]Q=VIt[/tex]

where [tex]i=current[/tex]

[tex]V=Voltage[/tex]

[tex]8.5\times 120\times t=2.4\times 4186\times 79[/tex]

[tex]t=778.10 s[/tex]

Final answer:

To determine the time required to boil the water in a coffee maker, calculate the power using P = IV, then the heat needed to raise the water temperature to its boiling point with Q = mcΔT, and finally divide Q by P to find the time.

Explanation:

The question deals with calculating how long it takes an automatic coffee maker to boil 2.4 kg of water using a resistive heating element. The initial temperature of the water is 21 °C, and we are given that the coffee maker is plugged into a 120 V electrical outlet with a current of 8.5 A. The specific heat capacity of water is 4186 J/kg°C. To find the time to boil the water, we'll first calculate the power delivered by the heating element using the relationship P = IV, where P is power, I is current, and V is voltage. Once power is known, we can use the equation Q = mcΔT to find the heat Q required to raise the temperature of water to its boiling point (100 °C), where m is water's mass, c is the specific heat capacity, and ΔT is the change in temperature. Finally, we'll divide quantified heat Q by the power P to find the time t: t = Q/P.

Answer:

U2 = 47.38m/s = initial velocity of B before impact

Explanation:

An example of the diagram is shown in the attached file because of missing angle of direction in the question

Mass A, B are mass of cars

A = 1965

B =1245

U1 = initial velocity of A = 52km/hr

U2 = initial velocity of B

V = common final velocity of two cars

BU2 = (A + B)*V sin ¤ ...eq1 y plane

AU1 = (A + B) *V cos ¤ ....equ 2plane

From equ 2

V = AU1/(A + B)*cos ¤

Substitute V into equation 1

We have

U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°

Substitute all parameters to get

U2 = (1965/1245)*52 * tan 30°

U2 = 47.38m/s

To find the corresponding velocity of car B just before impact, we can use the principle of conservation of momentum. By equating the total momentum before the collision to the total momentum after the collision, we can solve for the velocity of car B. Plugging in the given values and calculating, we can find the corresponding velocity of car B just before impact.

To solve this problem, we can use the principle of conservation of momentum. Since the two cars stick together after the collision, the total momentum before the collision is equal to the total momentum after the collision.

Let's define the velocity of car A just before impact as vA and the velocity of car B just before impact as vB.

Using the principle of conservation of momentum, we can write:

(mass of car A * velocity of car A) + (mass of car B * velocity of car B) = (total mass * common velocity)

Plugging in the given values, we have:

(1965 kg * 52 km/h) + (1245 kg * vB) = (3210 kg * common velocity)     (Note: Convert km/h to m/s)

Simplifying the equation, we get:

101820 kg * km/h + 1245 kg * vB = 3210 kg * common velocity

Now, we can solve for the velocity of car B just before impact:

vB = (3210 kg * common velocity - 101820 kg * km/h) / 1245 kg

By substituting the values and calculating, we can find the corresponding velocity of car B just before impact.

The speed of the metal ball in circular motion is calculated by using the equation for centripetal acceleration (v = √(rg)) and substituting the given values for radius and gravity. After calculation, the speed of the ball is determined to be approximately 5.1 m/s.

In this physics problem, we can calculate the speed of the metal ball by using the provided values and the concept of centripetal acceleration. Centripetal acceleration results from the force that makes the metal ball move in a circular pattern, which in this case is provided by the tension in the thread and gravity. In a scenario where the ball is in uniform circular motion, with the tension at an angle, forming a cone around the vertical axis, the component of tension providing the centripetal force equals the gravitational force or weight of the object.

A key equation that can be used here is: ac = v² /r, where 'ac' is the centripetal acceleration, 'v' is the velocity (or speed), and 'r' is the radius. Given that ac = g (acceleration due to gravity = 9.8 m/s²), we can modify the equation to: v = √(rg), where 'r' is the radius or the length of the thread (2.7 m).

Implementing these values into our equation we find that: v = √[(2.7 m)(9.8 m/s²)] = √26.46 = 5.1 m/s. Therefore, the speed of the metal ball when it is in circular motion is approximately 5.1 m/s.

https://brainly.com/question/14465119

#SPJ12

Answer

given,

mass of the car = 1600 Kg

velocity of the car = 21 m/s

time = 1.8 s

a) momentum = mass x velocity

P = 1600 x 21

P = 33600 kg.m/s

[tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]

[tex]Force = \dfrac{33600-0}{1.8}[/tex]

F = 18666.67 N

b) [tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]

[tex]Force = \dfrac{33600-0}{0.14}[/tex]

F = 240000 N

Answer:

a) [tex]T=1.02 s[/tex]

b) [tex]v=0.28\frac{m}{s}[/tex]

c) [tex]v=0.25\frac{m}{s}[/tex]

Explanation:

a) The period is defined as:

[tex]T=\frac{2\pi}{\omega}[/tex]

[tex]\omega[/tex] is the natural frequency of the system, in this case is given by:

[tex]\omega=\sqrt{\frac{k}{m}}\\\omega=\sqrt{\frac{9.5\frac{N}{m}}{250*10^{-3}kg}}\\\omega=6.16\frac{rad}{s}[/tex]

Now, we calculate the period:

[tex]T=\frac{2\pi}{6.16\frac{rad}{s}}\\T=1.02 s[/tex]

b) According to the law of conservation of energy, we have:

[tex]\frac{kx^2}{2}=\frac{mv^2}{2}\\v=\sqrt{\frac{kx^2}{m}}\\v=\sqrt{\frac{(9.5\frac{N}{m})(4.5*10^{-2}m)^2}{250*10^{-3}kg}}\\v=0.28\frac{m}{s}[/tex]

c) In this case, we have:

[tex]U_i=U_f+K_f\\\frac{kx_i^2}{2}=\frac{kx_f^2}{2}+\frac{mv^2}{2}\\v=\sqrt{\frac{k(x_i^2-x_f^2)}{m}}\\v=\sqrt{\frac{9.5\frac{N}{m}((4.5*10^{-2}m)^2-(2*10^{-2}m)^2)}{250*10^{-3}kg}}\\v=0.25\frac{m}{s}[/tex]

The maximum speed is; 0.28 m/s

The speed when it is located 2.0 cm from its equilibrium position is; 0.25 m/s

A) Formula for period is;

T = √(2π/ω)

where ω which is the natural angular frequency

ω = √(k/m)

we are given;

k = 9.5 N/m

m = 250 g = 0.25 kg

x = 4.5 cm = 0.045 m

Thus;

ω = √(9.5/0.25)

ω = 6.1 rad/s

Thus;

Period is; T = √(2π/6.1)

T = 1.02 s

B) From law of conservation of energy;

maximum speed is;

v = √(kx²/m)

v = √(9.5 * 0.045²/0.25)

v = 0.28 m/s

C) We are now given;

x₁ = 4.5cm = 0.045 m

x₂ = 2 cm = 0.02 m

Thus, speed is gotten from;

v = √((k(x₁ - x₂)/m)

v = √((9.5(0.045 - 0.02)/0.25)

v = 0.25 m/s

Read more about speed at; https://brainly.com/question/26114128

Answer:

Explanation:

Heres the correct and full question:

For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.

answer:

volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³

density of styrofoam=ρs=300kg/m³

mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg

weight of styrofoam = ws=(ms)g=360 x 9.8=3528N

consider number of contestants= n

toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)

since, styrofoam fully submerged into water, then bat force,

B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N

At equilibrium,

B - W - Ws = 0

12112.8 - 686n - 3528 = 0

[tex]n=\frac{12112.8-3528}{686}=12.5[/tex]

n=12 person(contestants)

Answer:

both have plasma membrane, ribosomes, cytoplasm, and DNA

Explanation:

Both eukaryotic and prokaryotic cells contain ribosomes, which are responsible for protein synthesis. Eukaryotic cells are more complex with a true nucleus, found in animals and plants, while prokaryotic cells, found in bacteria, are simpler and lack a true nucleus.

Option A is the correct answer. Both eukaryotic and prokaryotic cells contain ribosomes. The ribosomes in both types of cells serve the same function: protein synthesis, or assembly of amino acids to form proteins. Eukaryotic cells are those with a true nucleus and complex structures, found in animals, plants, fungi, and protists, while prokaryotic cells are smaller, simpler cells without a true nucleus, found in bacteria and archaea.

Unlike eukaryotic cells, prokaryotic cells don't contain a membrane-bound nucleus or mitochondria. A capsule is found in some bacteria, not all prokaryotes and no eukaryotes.

https://brainly.com/question/32100370

#SPJ2

Answer:

The total distance traveled by an object attached to a spring that is pulled to position x=A and then released is 4 A.

Explanation:

We have an small object attached to a relaxed spring whose opposite end is fixed. The spring rests on a frictionless surface. This means the only force acting on the object is the elastic  force of the spring, a conservative force, since the weight and the normal force compensate between them.

The initial position of the object is x=0 then is pulled to position x=A and released. After which it undergoes a simple harmonic motion with an amplitude A. From the position x=A to the equilibrium position in x=0 the object travels a distance A. From the equilibrium position x=0 to maximum negative position in x= -A the object travels again a distance A. Then to return to the original position the object should travel a distance 2 A in reverse direction.

In one full cycle of its motion the object travels a distance 4 A.

Answer:

P₂  = 370 kPa

Explanation:

Boyle's  law: States that the volume of a given mass of gas is inversely proportional to its pressure provided the temperature remains constant. It can be expressed mathematically as,

P₁V₁ = P₂V₂................ Equation 1

Where P₁ = Initial Pressure, V₁ = Initial volume, P₂ = Final Pressure, V₂ = Final Volume.

Making P₂ The subject of the equation above,

P₂ = P₁V₁/V₂..................... equation 2

Substituting these vaues into equation 2,

Where P₁ = 163 kPa = 163000 pa, V₁ =  10 L, V₂ = 4.4 L.

P₂ = 163000(10)/4.4

P₂ = 370454.55 Pa

P₂ = 370000 Pa

P₂  = 370 kPa To 3 significant figure.

Therefore Final pressure = 370 kPa

Final answer:

The maximum shear stress developed in the shaft is approximately 7.57 MPa. The angle of twist in the shaft at full power is approximately 3.00°.

Explanation:

To determine the maximum shear stress developed in the shaft, we can use the formula:

Shear stress (τ) = Torque (T) / Polar Moment of Inertia (J)

To find the torque, we can use the formula:

Torque (T) = Power (P) / Angular Velocity (ω)

We are given that the power is 2500 hp and the angular velocity is 1700 rpm. Converting these values to W and rad/s respectively, we have:

Power (P) = 2500 hp * 745.7 W/hp ≈ 1,864,250 W

Angular Velocity (ω) = 1700 rpm * 2π rad/minute ≈ 17897.37 rad/s

Substituting these values into the torque formula, we have:

Torque (T) = 1,864,250 W / 17897.37 rad/s ≈ 103.98 N*m

Next, we need to find the polar moment of inertia (J) of the shaft. The polar moment of inertia for a solid shaft can be calculated using the formula:

J = (π/2) * (outer diameter^4 - inner diameter^4)

Converting the diameter to meters, we have:

Diameter (d) = 8 in * 0.0254 m/in = 0.2032 m

Substituting this value into the polar moment of inertia formula, we have:

J = (π/2) * (0.2032^4 - (0.2032 - 2 * 3/8)^4)

= (π/2) * (0.2032^4 - 0.1926^4)

0.013743 m^4

Finally, we can calculate the maximum shear stress using the formula:

Shear stress (τ) = 103.98 N*m / 0.013743 m^4 ≈ 7.57 MPa

As for the angle of twist, we can use the formula:

Angle of twist (θ) = T * L / (G * Given that the length of the shaft is 100 ft and Young's modulus (G) for A-36 steel is approximately 200 GPa, we can calculate the angle of twist as follows:

Angle of twist (θ) = 103.98 N*m * 100 ft * 0.3048 m/ft / (200 GPa * 0.013743 m^4)

≈ 0.0524 radians or 3.00°

Answer:

Explanation: