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pshichka [43]
2 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]2 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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As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

here we know that

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d_i = -15 cm

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8 0
2 years ago
Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You
AleksAgata [21]

Answer:

\eta=0.5074\ or\ 50.74\%

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

  • density \rho=1\ g.mL^{-1}
  • specific heat c=4.186\ J.g^{-1}.K^{-1}
  • mass of coffee, m=200\times 1=200\ g
  • initial temperature of coffee, T_i=30^{\circ}C
  • final temperature of coffee, T_f=60^{\circ}C
  • power rating of oven, P=1100\ W
  • time taken to reach the final temperature, t=35\ s

<u>Heat released by the coffee to come to 60°C:</u>

Q=m.c.\Delta T

Q=200\times 4.186\times 30

Q=[tex]\eta=\frac{25116}{49500}\ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

E=P.t

E=1100\times 45

E=49500\ J

Now the efficiency:

\eta=\frac{Q}{E}

\eta=0.5074\ or\ 50.74\%

8 0
2 years ago
An ant moves towards the plane mirror with speed of 2 m/s &amp; the mirror is moved towards the ant with the same speed. What is
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  • Speed of mirror =v_b=2m/s

We know

\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}

\\ \sf\longmapsto V_{AB}=2-2

\\ \sf\longmapsto V_{AB}=0m/s

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2 years ago
Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo
Nutka1998 [239]
I made the drawing in the attached file.

I included two figures.

The upper figure shows the effect of:

- multiplying vector A times 1.5.
 It is drawn in red with dotted line.

- multiplying vector B times - 3 .
It is drawn in purple with dotted line.

In the lower figure you have the resultant vector: C = 1.5A - 3B.

The method is that you translate the tail of the vector -3B unitl the point of the vector 1,5A, preserving the angles.

Then you draw the arrow that joins the tail of 1,5A with the point of -3B after translation.

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7 0
2 years ago
Read 2 more answers
A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or
love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m  ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

f=\dfrac{nV}{4L}

f=\dfrac{3\times 344}{4\times 2}

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²)  x 1.5 x  10⁻³ x 100

T=1022.42 N

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2 years ago
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