How much momentum does a 10 kg object have when is moving 30 m/s

Which of the following is true concerning hurricanes?

densk [106]

Answer:

B.

Explanation:

B. is correct because every time usually when a hurricane hits it causes flooding causing multiple homes to be ruined. It is a known fact that usually hurricanes start close or in along with tornadoes, the water, meaning once they hit land they have some water to flood that land with.

C. is absolutely false hurricanes have VERY strong wind gusts.

D. is absolutely wrong they do alter landscapes by ripping trees and plants and houses out of the ground making the landscapes look different.

A. is wrong they tend to deposit and remove sediment evenly.

<em><u>~ LadyBrain</u></em>

6 0

3 years ago

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A projectile is fired at a velocity of 50 m/s straight into the air. What is the velocity of the projectile when the objects dis

Arisa [49]

Answer:

the final velocity of the object is 53.04 m/s.

Explanation:

Given;

initial velocity of the projectile, u = 50 m/s

displacement of the object, d = 16 m

let the final velocity of the object = v

Apply the following kinematic equation to determine the final velocity of the projectile.

v² = u² + 2gd

v² = 50² + (2 x 9.8 x 16)

v² = 2813.6

v = √2813.6

v = 53.04 m/s

Therefore, the final velocity of the object is 53.04 m/s.

8 0

3 years ago

PLEASE HELP what is the velocity of an airplane that travels 1,200 km in 2.2 hours

MA_775_DIABLO [31]

90.9 is the velocity of an airplane

7 0

3 years ago

Suppose that a car skids 15 meters if it is moving at 50 kilometers per hour Acceleration-Velocity-Position when the brakes are

julia-pushkina [17]

Explanation:

Below is an attachment containing the solution.

4 0

3 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

3 years ago