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4 years ago

8

A tennis player swings her 1000 g racket with a speed of 12 m/s. she hits a 60 g tennis ball that was approaching her at a speed

of 16 m/s. The ball rebounds at 42 m/s.
(a) How fast is her racket moving immediatley after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.

__________ m/s

(b) If the tennis ball and racket are in contatc for 10 ms, what is the average force that the racket exerts on the ball?

________ N

How does this compare to the gravitational force on the ball?

F avg / W ball=________

1 answer:

tensa zangetsu [6.8K]4 years ago

6 0

Answer:

should be the answer B

Explanation:

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A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What

nignag [31]

Answer:

speed of star is 1.37 × $10^{5}$ m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest = velocity of source / speed of light

so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

velocity of source star = 1.37 × $10^{5}$ m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

6 0

3 years ago

Why is it not necessary for radio telescope surfaces to be as smooth as a mirror?

Law Incorporation [45]

It doesn't on account of radio waves are longer than optical waves. Radio waves are a sort of electromagnetic radiation with wavelengths in the electromagnetic range longer than infrared light. These long waves are in the radio locale of the electromagnetic range.

8 0

3 years ago

A cart loaded with bricks has a total mass of 22.2 kg and is pulled at constant speed by a rope. The rope is inclined at 27.5 ◦

musickatia [10]

Answer:

W = 1.432 KJ

Explanation:

given,

mass = 22.2 Kg

angle of the rope = 27.5°

distance on the ground = 24 m

kinetic friction= μ = 0.32

acceleration due to gravity, g = 9.8 m/s²

Work done = ?

W = F d cosθ

a = 0 because it is moving with constant speed

equating all the forces acting in x direction

F cosθ = F friction = μN

equating all the forces acting in y direction

F sinθ + N -mg =0

now,

N = mg - F sinθ

putting value of N

F cosθ = μ mg -μ F sinθ

F (cosθ + μsinθ ) = μ mg

$F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}$

$F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}$

F =67.28 N

now,

W=F d cosθ

W =67.28 x 24 x cos(27.5)

W =1432.27 J

W = 1.432 KJ

7 0

3 years ago

Read 2 more answers

Please help with this question.

EastWind [94]

Answer:

41.16 Joules

Explanation:

Potential energy at a given instant is a function of mass and height of an object. The formula is

$E_p = mgh = 2.80kg\cdot 9.8\frac{m}{s^2}\cdot 1.50m = 41.16 J$

4 0

3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

3 years ago