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Over [174]
3 years ago
7

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

orizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

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A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,
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Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

Q_{water}=Q_{methanol}

where:

Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})

Now replacing:

0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]

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what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?​
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Answer:

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