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bogdanovich [222]
3 years ago
10

A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What is the velocity of the bowling ball?

Physics
1 answer:
KIM [24]3 years ago
7 0

Answer:

6.19m/s

Explanation:

Given parameters:

Mass of bowling ball  = 35.2kg

Momentum  = 218kgm/s

Unknown:

Velocity of the bowling ball  = ?

Solution:

Momentum is the quantity of motion a body possess;

    Momentum  = mass x velocity

   218  = 35.2 x velocity

     Velocity  = 6.19m/s

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kodGreya [7K]

Answer:

\Delta H=687.4 J

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J

\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J

Therefore, the required heat is:

\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

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2 years ago
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Answer:

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5 0
3 years ago
Which of the following is true about Viscosity of liquids:
White raven [17]
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A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all
timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, a=5\times 10^{-4}\ m^2

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

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P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}

P=1.4\times 10^5\ Pa

We know that the atmospheric pressure is given by :

P_o=1.013\times 10^5\ Pa

So, the pressure is given by :

p=P+P_o

p=1.4\times 10^5+1.013\times 10^5

p=2.41\times 10^5\ Pa

Hence, this is the required solution.

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Answer:

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