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AleksandrR [38]
3 years ago
13

A charge of 4.5x10^-5 C is placed in an electric field with a strength of 2.0x10^4 N/C. What is the electric force acting on the

charge?
Physics
1 answer:
Alex3 years ago
6 0

Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

F=qE

where

q is the magnitude of the charge

E is the strength of the electric field

In this problem, we have

q=4.5\cdot 10^{-5}C is the charge

E=2.0\cdot 10^4 N/C is the strength of the electric field

Substituting into the equation, we find

F=(4.5\cdot 10^{-5}C)(2.0\cdot 10^4 N/C)=0.9 N

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2 years ago
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

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Learn more about height of projectiles here: brainly.com/question/10008919

6 0
2 years ago
Why is acceleration not constant near the speed of light
dmitriy555 [2]

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2 years ago
In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
galina1969 [7]

Answer:

W_x = 0.9156\ c

Explanation:

given,

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for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

5 0
2 years ago
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