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Alenkasestr [34]
3 years ago
9

What is the measure of an average kinetic energy??

Physics
1 answer:
makvit [3.9K]3 years ago
7 0
Temperature is the energy
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The maximum distance particles of the medium move when a wave passes through them is wave.
joja [24]

Answer:

amplitude

Explanation:

6 0
2 years ago
A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t
vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be m kg.

Given:

Initial velocity of the car is, u=25\ m/s

As the car stops, final velocity of the car is, v=0\ m/s

Change in momentum is, \Delta p=39400

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg

Therefore, the mass of the car is 1576 kg.

4 0
3 years ago
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what
Katarina [22]

Answer:

Hiii how are you <u>doing?</u><u>?</u><u>I </u><u>don't</u><u> </u><u>understand</u><u> </u><u>that</u>

3 0
2 years ago
Lightning a. occurs when a positively charged cloud base induces a negative charge on the Earth's surface. b. travels most often
mart [117]

Answer:

c. may be accompanied by the sound of explosively expanding hot air, called thunder.

Explanation:

Lightning is a discharge which is due to the reaction between oppositely charged charges in the clouds, or between clouds base and the Earth surface.

The motion of the cloud causes charging of clouds by friction, thus the reaction between opposite charges (jumping of charges from one cloud to another) in the cloud can lead to lightning. Also, oftentimes the bottom of a cloud is negatively charged so that this is attracted to the positive charge on the earth surface. Thus leading to a discharge called lightning.

Thus in the given question, the appropriate option is C. This implies that, lightning may be accompanied by the sound of explosively expanding hot air, called thunder.

8 0
3 years ago
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