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lord [1]
3 years ago
9

Help me plz with this

Physics
1 answer:
sveticcg [70]3 years ago
5 0
I believe it is green at 1,550
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The fastest speed ever measured for a tennis ball served by a player was 263 km/h. The distance of the tennis court from one end
Paha777 [63]

Answer:

0.3257 seconds

39.67 m/s

Explanation:

Speed = 263 km/h

Converting to m/s

1\ km/h=\frac{1}{3.6}\ m/s

263\ km/h=263\times \frac{1}{3.6}\ m/s=\frac{1315}{18}\ m/s

Distance to travel = 23.8 m

Time = Distance / Speed

Time=\frac{23.8}{\frac{1315}{18}}=0.3257\ s

The time taken to go from one end of the court to the other is 0.3257 seconds

Time = 0.6 s

Speed = Distance / Time

Speed=\frac{23.8}{0.6}=39.67\ m/s

The speed of the tennis ball is 39.67 m/s

7 0
3 years ago
Which physical property is best measured using only a balance?
Step2247 [10]
The answer is C-Mass
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This is a cell, which is the basic unit of all life. All organs in human bodies are made of cells and require oxygen to survive.
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The two systems that work together to deliver oxygen are D, respiratory and cardiovascular
5 0
3 years ago
Work is done on an object when an applied force causes the object to move in the same direction as the force.
Lelu [443]

Answer:yes

Explanation:

Work is done on an object when an applied force causes the object to move in the same direction as the force

3 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

6 0
3 years ago
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