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Oduvanchick [21]
3 years ago
10

The charge that passes a cross-sectional area A=10-4 m2 varies with time according

Physics
1 answer:
Anarel [89]3 years ago
6 0

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

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Winston stands on the edge of a building's flat roof, 12 m above the ground, and throws a 147.0-g baseball straight down. the ba
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A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p
Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s

It takes 2.84403 seconds to reach the highest point

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

6 0
2 years ago
What is the atomic number of this element?
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5 0
3 years ago
Read 2 more answers
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10âˆ
AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, F=3.33\times 10^{-21}\ N

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

F=k\dfrac{q^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}

q=1.21\times 10^{-16}\ C

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

n=\dfrac{q}{e}

n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0
3 years ago
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