What is responsible for creating distinct weather patterns?

Witch sentence states Newton third law

Dominik [7]

Answer:

For every action, there is an equal and opposite reaction.

Explanation:

Physics helps alot lol

7 0

3 years ago

. A book is moved once around the perimeter of a table of dimensions 2.00 m by 3.00 m. If the book ends up at its initial positi

Lynna [10]

Answer:

10

Explanation:

displacement would be 10 because knowledge

5 0

2 years ago

A 500N car is pulled up to 20m plant to a flat from 5m above the ground by an effort of 150N parallel to the plank what is the V

Stolb23 [73]

Answer:

Explanation:

VR = Velocity ratio = Distance moved by effort/Distance moved by load

Given:

Distance moved by effort = 20m

Distance moved by Load = 5m

VR = 20/5

VR = 4

Hence the velocity ratio of the car is 4

6 0

2 years ago

¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?

VladimirAG [237]

Answer:

I will answer this in English, we can translate it to:

Why if you charge a mate by an amount of time you are not doing work?

This happens because work is defined as the displacement done by a force:

W = d*F

where W is work, d is the distance, and F is the force.

This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.

4 0

3 years ago

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within

zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0

2 years ago