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iren2701 [21]
4 years ago
7

The mass of a regulation tennis ball is 57 g. The ball is in contact with the tennis racket for 30 ms. The ball was served at 73

.14 m/s. If the opponent returned the serve with a speed of 55 m/s, what force did he exert on the ball, assuming only horizontal motion?
Physics
1 answer:
yaroslaw [1]4 years ago
8 0

Answer:

Correct answer:  F₂ = 104.5 N

Explanation:

Given:

m = 57 g = 57 · 10⁻³ kg

Δt = 30 ms = 30 · 10 ⁻³ seconds

V₁ = 73.14 m/s   service speed

V₂ = 55 m/s  returned speed

M = m · V  Momentum or Impulse

You forgot to indicate what time the ball contact when returning.

We will assume that the time is the same Δt = 30 ms = 30 10 ⁻³ seconds.

The formula for calculating force is according to Newton's second law is:

F = ΔM / Δt = m · ΔV / Δt

Force during service is:

F₁ = 57 · 10⁻³ · 73.14 / 30 · 10 ⁻³ = 138.97 N

F₁ = 138.97 N

Returned force:

F₂ =  57 · 10⁻³ · 55 / 30 · 10 ⁻³ = 104.5 N

F₂ =  104.5 N

God is with you!!!

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3 years ago
A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
4 years ago
The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro
irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}

f = 34.047 cm = 0.3404 m

Power of the lens = P

P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter

6 0
3 years ago
An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

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