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pshichka [43]
4 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]4 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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1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.
yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, M=5.3\times 10^3\ A/m

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

M=\dfrac{\mu}{V}

\mu=M\times \pi r^2\times l

Where

r is the radius of rod, r = 0.0075 m

\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m

\mu=0.0683\ J/T

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

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