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Sidana [21]
3 years ago
10

What name is given to the rate of flow of electric charge?

Physics
1 answer:
RSB [31]3 years ago
3 0

Answer:

<h2>Electric charge</h2>

Explanation:

The rate of the flow of electric charge is known as electric current. <u>By convention, the direction of electric current is always the direction of net flow of positive charge.</u>

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#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

\\ \sf\longmapsto F_L=\mu sN

u s is coefficient of static friction and N is normal reaction

Or

\\ \sf\longmapsto F_L=\mu smg

  • As N=mg
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<span>Barium-139 is the correct answer.</span>
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A bowl-shaped depression formed by a mountain glacier is termed a(n) ____________.
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Name the elements from which nichrome can be made.​
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7 0
3 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 25
Ivanshal [37]

Answer:

a) v(2\,s) = 5.9\,\frac{m}{s}, v(4\,s) = -13.7\,\frac{m}{s}, b) t = 2.602\,s, c) h(2.602\,s) = 35.176\,m, d) t = 5.281\,s, e) v(2.602\,s) = -26.254\,\frac{m}{s}

Explanation:

a) The velocity function is determined by deriving the position function in time:

v(t) = 25.5-9.8\cdot t

Velocities after 2 seconds and 4 seconds are, respectively:

v(2\,s) = 5.9\,\frac{m}{s}

v(4\,s) = -13.7\,\frac{m}{s}

b) The maximum height is reached when velocity is equal to zero:

25.5-9.8\cdot t = 0

The time when the projectile reaches the maximum height:

t = 2.602\,s

c) The maximum height is:

h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}

h(2.602\,s) = 35.176\,m

d) The projectile hits the ground when height is equal to zero:

-4.9\cdot t^{2}+25.5\cdot t + 2 =0

The roots of the second order polynomial are presented below:

t_{1} \approx 5.281\,s

t_{2} \approx -0.077\,s

The first one is the only reasonable solution in physical terms.

t = 5.281\,s

e) The velocity of the projectile when it hits the ground is:

v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)

v(2.602\,s) = -26.254\,\frac{m}{s}

4 0
3 years ago
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