Describe the motion of an object that has an acceleration of 0 mi./s squared

Jobisdone [24]

$\text{Acceleration}=\frac{d^2}{dt^2}\text{displacement}$
So the dimensions of acceleration is $LT^{-2}$
Any answer that comes under that definition is correct.

6 0

3 years ago

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When bouncing a ball, the bouncing motion results in the ball ____________.

yan [13]

Answer:

B) changing position

Explanation:

When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, due to multiple bounce the position of the ball changes.

Thus, When bouncing a ball, the bouncing motion results in the ball changing position.

6 0

3 years ago

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PLSSSS ANSWER!!!!!!!!!!!!!

Taya2010 [7]

Answer:

d

Explanation:

7 0

2 years ago

A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan

Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

$E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C$

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

$E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C$

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0

2 years ago

What are the strengths and limitations of the doppler method?

GrogVix [38]

Answer:

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

Explanation:

Strengths: yield planet's size

Limitations: few planets have the necessary orbital alignment to be detectable

4 0

2 years ago