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2 years ago

How tall is Mount Everest?

Physics

2 answers:

nikklg [1K]2 years ago

6 0

Answer: 29,035 feet

Explanation:

White raven [17]2 years ago

3 0

Answer:

29,032′

Explanation:

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How far out from the sun is the rock line?

Vesna [10]

Answer:

5 million miles

Explanation:

8 0

2 years ago

A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what

Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

$P=m*v\\$

where:

m = mass [kg]

v = velocity [m/s]

$P =20*10\\P =200 [kg*m/s]$

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

$v = P/m\\v = 200/5\\v = 40 [m/s]$

3 0

2 years ago

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago

What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 0

3 years ago

NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25

poizon [28]

22. reduction
25. Le Chatelier's principle

8 0

3 years ago