Weight and mass are interchangeable terms. True or False

2 answers:

8 0

False
weight (N-force) = mass(kg-matter) x g

strictly in Physics, both are not interchangeable.
in normal life, weight (kg) actally means mass in Physics.

Nonamiya [84]3 years ago

5 0

It's false. Mass is a way of measuring how much matter an object contains, where as weight measures how hard gravity is pulling on an object. While on earth, these are typically interchangeable. However, if you were to go to Mars, your mass would stay the same, but the weight will be different. This is because you still contain the same amount of matter, but the gravity's pull will be different because the moon has a different gravitational pull than the earth. Hope this helps!

You might be interested in

An icicle falls off of a skyscraper from rest and falls for 34 seconds. How fast will that icicle be moving after that time? Sho

goblinko [34]

Answer:

<em>The icicle will be moving at 333.54 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.81 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The icicle falls from rest for 34 seconds. We need to find the speed after that time:

vf = 9.81*34

vf = 333.54 m/s

The icicle will be moving at 333.54 m/s

7 0

3 years ago

On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s

omeli [17]

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)

h₀ - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

$h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m$