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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

2 answers:

EleoNora [17]3 years ago

8 0

The basketball is gaining kinetic and losing potential energy is the answer

Radda [10]3 years ago

4 0

Answer: Option (A) is the correct answer.

Explanation:

Kinetic energy is the energy obtained by an object due to the motion of its particles. Whereas potential energy is the energy obtained by an object due to its position.

At point A basket ball is at rest so, it will have maximum kinetic energy.

When the basket ball bounces from position A to position B then potential energy will convert into kinetic energy.

Thus, we can conclude that as the basketball bounces up from A to B it is gaining kinetic energy and loosing potential energy as it bounces up.

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

Darya [45]

<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>

5 0

3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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2 years ago

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3 years ago

What's an acrostic poem for climate each line has to be written in a full sentence?

Mrac [35]

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3 years ago

How fast did Newton decide a cannon ball shot out in order to escape Earth’s gravity?

arsen [322]

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3 years ago