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3 years ago

Gravitational force depends on the mass of an object.

Physics

1 answer:

vampirchik [111]3 years ago

4 0

Answer:

sun, earth, capitol building, human, atom

Explanation:

Gravity is directly proportional to the size of an object. Therefore, the object with the most mass will have the greatest force of gravity. We know this from the equation. $F = G\frac{m_1m_2}{r^2}$

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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3 years ago

What is the area called where the particles are spread apart?

Ostrovityanka [42]

Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively

Explanation:

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3 years ago

What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

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3 years ago

What are some evidences of early peoples concern for teeth

Alexandra [31]

Replacement teeth were made even back then out of primitive materials, and pig hair was used like floss

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4 years ago

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa

Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
This force is not a new force, is just the net force aiming to the center of the circle.
In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
So, we can write the following expression:

$F_{cent} = F_{n} - F_{g} (1)$

It can be showed that the centripetal force is related to the speed by the following expression:
$F_{cent} = m*\frac{v^{2}}{r} (2)$

The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
Replacing (2) in (1), and solving for Fn, we get:

$F_{n} = m*\frac{v^{2} }{r} + m*g (3)$

Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

$F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

Replacing Fg by its value, simplifying, and solving for v, we get:

$v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$

3 0

3 years ago