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Nana76 [90]
3 years ago
11

4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?

Physics
1 answer:
pentagon [3]3 years ago
5 0

The electric field strength is 2.16\cdot 10^{26} N/C

Explanation:

The strength of the electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge at which the field strength is calculated

For the charge in the problem, we have:

q=4.7 cC = 0.047 C is the charge

r=1.4 nm = 1.4\cdot 10^{-9} m

Therefore, the electric field strength is

E=(8.99\cdot 10^9)\frac{0.047}{(1.4\cdot 10^{-9})^2}=2.16\cdot 10^{26} N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

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A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is r
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Answer:

J = 1800 kg-m/s

Explanation:

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