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anygoal [31]
3 years ago
6

Which equation is used to calculate the weight of an object on a planet?

Physics
2 answers:
kondaur [170]3 years ago
8 0

W=mg

Explanation:

the weight of an object on a planet is given by the law:

W=mg

where

m is the mass of the object

g is the value of the gravitational acceleration on the planet

While the mass of the object, m, depends only on the properties of the object (so, it is the same on every planet), the weight W, depends on the value of g, which is different from planet to planet, therefore the weight of an object changes if we move it to another planet.

Kamila [148]3 years ago
6 0

In this case to find the weight of an object you must use the formula.

W = mg

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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
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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 weight on the surface of Earth of an object of mass 2.00 kilograms?
Anni [7]

Answer:

D

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

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2 years ago
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The exosphere is the layer of the atmosphere
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The exosphere is the layer of the atmosphere "Where gas molecules can be exchanged between Earth's atmosphere and outer space." Thus, the answer would be C.
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A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh
melisa1 [442]

Answer: 1.2m/s^2

Explanation: the force exerted on the car is 900N upwards

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According to Newton's third law acceleration is proportional to force

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3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

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