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Zanzabum
3 years ago
15

Current is the movement of positive charges called electrons

Physics
2 answers:
enot [183]3 years ago
8 0

Answer:yes

Explanation:current is the movement of electrons along a conductor

GaryK [48]3 years ago
6 0

Answer:what

Yeah

Explanation:

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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
astra-53 [7]

Answer:

They all hit at the same time

Explanation:

Let the time of flight is T.

The maximum height is H and the horizontal range is R.

The formula for the time of flight is

T=\frac{2uSin\theta }{g} ..... (1)

Te formula for the maximum height is

H=\frac{u^{2}Sin^{2}\theta }{2g}    .... (2)

From equation (1) and (2), we get

\frac{T^{2}}{H}=\frac{\frac{4u^{2}Sin^{2}\theta }{g^{2}}}{\frac{u^{2}Sin^{2}\theta }{2g}}    

\frac{T^{2}}{H}=\frac{8}{g}

T=\sqrt{\frac{8H}{g}}

here, we observe that the time of flight depends on the maximum height and according to the question, the maximum height for all the three balls is same so the time of flight of all the three balls is also same.

8 0
3 years ago
Explain why gaps are left between railway lines​
Anuta_ua [19.1K]

Answer:

to prevent the derailing of the train due to thermal expansion.

3 0
3 years ago
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A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car
dangina [55]
Her magnitude of deceleration on the ice would be 15.126m/s
5 0
3 years ago
State 7 branches of physics
erma4kov [3.2K]

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mechanics

optics

sound

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nuclear physics

heat and temperature

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are branches of physics

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3 years ago
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A car has a distance between axles of 6.2 m and a center of mass located 2.2 m from the front axle. What is the
NARA [144]

Answer:

The correct option is  C

Explanation:

From the question we are told that

    The distance between axles of the car  is  d_a  =  6.2

     The position of the car's center of mass is  D  = 2.2 \ m  \ in \ front \  of  \  the \  axel

Now  we can evaluate the distance of the center of mass to the rear axle as follows

        k  =  d_a  -  D

substituting values

      k  =  6.2 -  2.2

      k  = 4 \ m

assuming the car is   at equilibrium,  taking moment about the center of mass

          W_{front \  axle} *  D  -  W_{rear \  axle} *  k  = 0

=>        W_{front \  axle} *  D     =  W_{rear \  axle} *  k

=>       \frac{W_{front \  axle}}{W_{rear \  axle} }       =    \frac{k}{D}

substituting values

             \frac{W_{front \  axle}}{W_{rear \  axle} }       =    \frac{4}{2.2}

           \frac{W_{front \  axle}}{W_{rear \  axle} }       =    1.81       Note [W_{front \  axle} is the front axle weight  and  

                                                      W_{rear \  axle} is the rear axle weight  ]

6 0
3 years ago
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