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shutvik [7]
3 years ago
8

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

Answer:

\frac{inertia_B}{inertia_A}=9

Explanation:

First of all, let's remind that:

- The kinetic energy of an object is given by K=\frac{1}{2}mv^2, where m is the mass and v is the speed

- The momentum of an object is given by p=mv

- The inertia of an object is proportional to its mass, so we can write I=km, where k just indicates a constant of proportionality

In this problem, we have:

- K_A = K_B (the two objects have same kinetic energy)

- p_A = 3 p_B (A has three times the momentum of B)

Re-writing both equation we have:

\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B

If we divide first equation by second one we get

v_A = 3 v_B

And if we substitute it into the first equation we get

m_A (3 v_B)^2 = m_B v_B^2\\9 m_A v_B^2 = m_B v_B^2\\m_B = 9 m_A

So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

\frac{I_B}{I_A}=\frac{km_B}{km_A}=\frac{9m_A}{m_A}=9

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Yuliya22 [10]

Answer:

The change in momentum of both objects is the same but in opposite direction.

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momentums of each glider. The momentum of the system is conserved if no external force is acting on the objects (as in this case). That means that the initial momentum of the system is equal to the final momentum of the system.

The momentum of each glider is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass of the glider.

v = velocity.

The momentum of the system for glider A and B can be calculated as follows:

initial momentum = mA · vA + mB · vB

Where:

mA and vA = mass and velocity of glider A

mB and vB = mass and velocity of glider B

Initially, glider B is at rest so that vB = 0. Then, the initial momentum of the system is:

initial momentum = mA · vA

The final momentum of the system is calculated as follows:

final momentum = mA · vA´ + mB · vB´

Where vA´ and vB´ are the final velocities of glider A and B respectively.

We know that mB = 4mA and that vA´ is negative. The the final momentum will be:

final momentum = -mA · vA´ + 4mA · vB´

Since initial momentum = final momentum:

mA · vA = -mA · vA´ + 4mA · vB´

mA · vA + mA · vA´ = 4mA · vB´

<u>vA + vA´ = 4 vB´</u>

<u />

The change in momentum of glider A (ΔpA) is calculated as follows:

ΔpA = final momentum - initial momentum

ΔpA =  -mA · vA´ - mA · vA = -mA (vA + vA´) = -4mA · vB´

The change in momentum of glider B (ΔpB) is calculated as follows:

ΔpB = final momentum - initial momentum

ΔpB = 4mA · vB´ - 0 = 4mA · vB´

Then, the change in momentum of both objects is the same but in opposite direction. That´s why the momentum is conserved.

4 0
3 years ago
I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
lianna [129]

Answer:

a. 250kg I think it's the right answer. hope it helps:)

7 0
3 years ago
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We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat
Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)

Here, mw is the mass of water

Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal

Now, you need to calculate the heat released by the steam:

Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal

The work done by this engine is the difference between both heats:

W=Q_{1}-Q_{2}=62760-61960=800cal

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A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s
Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

            Kinetic Energy, KE = \frac{1}{2} mv^2

Therefore,    KE = \frac{62 X  5.48 X 5.48}{2}

KE = 930.94 Joules

Ice melting lateral heat is  334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0
3 years ago
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