I appreciate the help on this physics question

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

yaroslaw [1]

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!!!

8 0

3 years ago

While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b

Citrus2011 [14]

Answer:

$\theta = 66.7 degree$

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

$F_n = mg + Fsin\theta$

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

$Fcos\theta = \mu F_n$

$F cos\theta = \mu (mg + F sin\theta)$

$F(cos\theta - \mu sin\theta) = \mu mg$

$F = \frac{\mu mg}{cos\theta - \mu sin\theta}$

so here we can say

$cos\theta - \mu sin\theta > 0$

$tan\theta = \frac{1}{\mu}$

$\theta = tan^{-1}\frac{1}{\mu}$

$\theta = tan^{-1}(\frac{1}{0.43})$

$\theta = 66.7 degree$

6 0

3 years ago

1. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is the

ANTONII [103]

Answer:

43,555 W

Explanation:

Since the speed is constant, this means that the force applied to lift the equipment is equal to the weight of the equipment. So we can write:

$F=mg$

where

m = 1000 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity

So,

$F=(1000)(9.8)=9800 N$

Then the work done in lifting the equipment is:

$W=Fd$

where

d = 200 m is the displacement of the equpment

Substituting,

$W=(9800)(200)=1.96\cdot 10^6 J$

Finally, the power used to lift the equipment is the ratio between the work done and time taken:

$P=\frac{W}{t}$

where

$W=1.96\cdot 10^6 J$

t = 45 s is the time taken

Solving,

$P=\frac{1.96\cdot 10^6}{45}=43,555 W$

8 0

3 years ago

When the chemical name of a compound is being written, the subscripts will determine the symbols. elements. prefixes. suffixes.W

Xelga [282]

Answer:

C. Prefixes

Explanation:

I took the quiz on edge

5 0

3 years ago

Read 2 more answers

Hãy nêu sự tương tự giữa công của lực điện trong trường hợp này với công của trọng lực.

statuscvo [17]

Answer:

tốt ý bạn là gì?

Explanation:

6 0

3 years ago