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topjm [15]
3 years ago
15

How does mass affect weight?​

Physics
1 answer:
devlian [24]3 years ago
5 0
Answer: Mass affects the weight of an object with the effects of gravity.

Weight is the measure of the force of gravity on an object's mass, while mass is the measure of how much matter there is in an object.
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Time period of a simple pendulum is measured at Karachi. What change will occur in the time period, if it is measured on mount e
Reil [10]
The period of a pendulum is given by T=2 \pi  \sqrt{ \frac{l}{g} }
since Karachi is near sea level, g is larger than it is on Mt. Everest.  That means the pendulum will have a larger period on Mt. Everest than it would in Karachi.
5 0
3 years ago
What's an easy way to create an interference pattern of waves?
igor_vitrenko [27]

Answer:

B

Explanation:

5 0
3 years ago
Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7
Norma-Jean [14]

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6


Therefore, v= 3.5 m/s.

7 0
3 years ago
Read 2 more answers
What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman = \frac{95.08}{100 } = 0.9508

The density of the woman is calculate as;

Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman  = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0
3 years ago
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