What is the acceleration that earth experiences due to your gravitational pull?

Physics

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

The acceleration that earth experiences due to gravitational pull is = 9.81 $\frac{m}{s^{2} }$ .

Explanation:

The acceleration that earth experiences due to gravitational pull is called the acceleration due to gravity. its value is 9.81 and its unit is $\frac{m}{s^{2} }$ .

When the object move upwards than in that case the earth gravitational force pulls down the body.

The formula of force due to gravity on the body is given as

F = mg

where g = acceleration due to gravity.

Due to this acceleration the body falls upon the surface of the earth.

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Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev

Yanka [14]

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

6 0

2 years ago

A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.

Pavlova-9 [17]

Answer:

v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

ρ g y₁ = ½ ρ v₂²

v₂ = $\sqrt {2g \ y_1}$