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BARSIC [14]
3 years ago
7

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

n the string makes a 30 angle with the vertical, what is the net charge on the ball

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer: Here is the complete question:

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?

Answer: The charge on the ball is 5.71 × 10^-4 C

Explanation:

Please see the attachments below

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Which wave characteristic has the unit hertz??
kotegsom [21]
Frequency has the unit hertz.
8 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
2 years ago
A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of
docker41 [41]

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

                                           =2 feet

                                           =2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

                     Distance Jumped=2+2

                     Distance Jumped=2*2

                                                   =4 feet

Similarly after 7 jumps

                    Distance Jumped=2+2+......+2

                    Distance Jumped=2*7

                                                 =14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

              Distance Jumped=(Distance Jumped after 7 jumps)+3

                                           =14+3

                                           =17 feet

The frog takes 8 jumps to reach top of well                

7 0
3 years ago
How do very high density objects appear in an ultrasound?
evablogger [386]
<span>Density is entirely unrelated to an object's size. It is a property of a given</span>
7 0
3 years ago
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
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