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nadezda [96]
3 years ago
7

PLEASE HELP An object is thrown upward with an initial velocity of 32.1 m/s. What is its velocity in 4.0 s? (use g = 9.81 m/s2).

Please include step by step process. -7.14 11.2 7.14 -4.6
Physics
1 answer:
grigory [225]3 years ago
6 0

Answer:

-7.14

Explanation:

According to equation of motion

v = u + at

where

v is the final velocity at any time t

u is the initial velocity

a is the acceleration

and t is the time

___________________________________________

Given

An object is thrown upward it means

its initial velocity is in upward direction

but acceleration due to gravity is in downward direction which will cause to decrease the velocity of object.

Intial velocity u = 31.1 m/s

final velocity v at 4 second we have to find.

t = 4 second

a = g = 9.81 m/s2 (it is acting in opposite direction of motion hence its sign will be negative ).

Thus

a = -  9.81 m/s2

using the above values in v = u + at

v = 32.1 - 9.81*4

v = 32.1 - 39.24

v = -7.14

Thus, correct option is -7.14.

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c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

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Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

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By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

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which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

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A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

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So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

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c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

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Gauss theorem this time becomes

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q=-17.0\mu C=-17.0\cdot 10^{-6}C

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Learn more about electric field:

brainly.com/question/8960054

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