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mamaluj [8]
3 years ago
14

A force of 210 N is applied to an object and the object accelerates at 14m/s2. Determine the mass of the object in kg.

Physics
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

15 kg

Explanation:

F = m*a

F/a = m

210/14 = 15 kg

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Why do cells undergo mitosis?
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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

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}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

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

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

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

q=0

So we have

\int E dS=0

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

The area of the surface is

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}

The charge included within the sphere in this case is the charge on the paint layer, therefore

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

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

where the negative sign means the direction of the field is inward, since the charge is negative.

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:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

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

And the charge included within the sphere is again the charge on the paint layer,

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

Therefore, 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})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

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5 0
3 years ago
I need help me with my question
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The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.

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  • The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
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Thus, we can conclude that the option D is correct.

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2 years ago
Classify the following situations into contact and non-contact forces.
NARA [144]

Answer: Contact force

a. Applying break in a vehicle.

d. The speed of ball rolling on ground is reduced

Non contact force

b. A coconut falling from a coconut tree.

c. The planets revolving around the sun.

Explanation:

The contact force is the force which exerts when one object or entity comes in contact with other object or entity. For example, on application of break the vehicle stops, the force is applied on the breaks to stop the vehicle. The ball rolling on the ground the speed reduces so the application of force on the ground also reduces.

The non contact force is the force one object exerts on the other without coming in direct contact with the other object. The force exerted by one object on other due to gravity is a non contact force. The coconut falling on the ground and planets revolving around the sun are examples of non contact force due to gravity.

8 0
3 years ago
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