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pshichka [43]
3 years ago
6

In being served, a tennis ball is accelerated from rest to a speed of 42.1 m/s. The average power generated during the serve is

2920 W. Assuming that the acceleration of the ball is constant during the serve, find the force acting on the ball.
Physics
1 answer:
hoa [83]3 years ago
8 0

Answer:

Force F = 69.35 N

Explanation:

given data

Ball Initial speed u = 0

Ball Final speed v = 42.1 m/s

average power generate = 2920 W

solution

Power generate is express as

P= \frac{W}{t}   ..............1

here W is work done and t is time

and work w = F × d

so

P= \frac{Fd}{t}  

and we know speed v = \frac{d}{t}  

so here

Power P = F × v

put here value and we get force

Force F = \frac{2920}{42.1}

Force F = 69.35 N

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If the temperature is increased the particles gain more kinetic energy or vibrate faster. This means that they move faster and take more space.

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A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache
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Answer:

F_a=1470\ N

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

\displaystyle F_{r2}-T=0

The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

8 0
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The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

<h3>De Broglie wavelength:</h3>

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

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λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

Learn more about de Broglie wavelength on

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Nothing, ATOMS lose or gain electrons to become IONS 
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If you double the diameter of a telescope, you double its light-collecting area.
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Answer:true

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