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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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3 years ago
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s.
8090 [49]
When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula
v^2 = u^2 - 2gh
where 
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.

Therefore
0 = 1.2^2 - 2*9.8*h
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5 0
3 years ago
A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

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now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

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now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

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4 years ago
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