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3 years ago

A value that describes how heavy an object is and is related to the force of gravity is

Physics

2 answers:

Slav-nsk [51]3 years ago

6 0

The heaviness of an object with the relation to force of gravity, is described by the ‘weight’

Explanation:

The heaviness of the object is referred to the mass of the object. When the mass of the object is concerned it is taken in space or vacuum where force of gravity do not exist.

Therefore, to value the actual heaviness of the object, mass is multiplied by the component of force of gravity, is termed as weight.

$\bold{\text { Weight }=\text { Mass } \times \text { Accleration due to gravity }}$

Natasha_Volkova [10]3 years ago

3 0

Answer:

mass

Explanation:

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3 years ago

A block of mass 1.5 hangs at the of end of a weight cord suspended from the ceiling.what is the tension in the cord, and with wh

Len [333]

The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

<h3>What is the tension in the cord?</h3>

The tension in the cord is calculated as follows;

T = ma + mg

where;

a is the acceleration of the block
g is acceleration due to gravity
m is mass of the block

T = m(a + g)

T = 1.5(a + 9.8)

T = 1.5a + 14.7

Thus, the tension in the cord is (1.5a + 14.7) N.

If the block is at rest, the tension is 14.7 N.

<h3>Force of the force</h3>

The force with which the cord pulls is equal to the tension in the cord

F = T = m(a + g)

F = (1.5a + 14.7) N

If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.

Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

Learn more about tension here: brainly.com/question/187404

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1 year ago

The Earth is constantly spinning on its axis, like you might spin a basketball on your finger. It is this spinning of the Earth

mash [69]

Answer:

The spinning of the earth around its own axis causes day and night.

Explanation:

Earth has two types of motions. It spins around its own axis causing day and night every 12 hours and completes a rotation in 23.93 hours that make a full day. The part of the earth that faces sunlight during spinning experiences day and the other part has night. It also rotates around the sun and completes one rotation in 365 days that makes a year.

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3 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)