Scales for weight
Any beaker to measure the volume of liquid displaced
Answer:
165.77J
Explanation:
M₁ = 0.107kg
u₁ = 300m/s
m₂ = 3kg
u₂ = 0
v =
m₁u₁ + m₂u₂ = (m₁ + m₂)V
(0.107*300) + 0 = (0.107 + 3)V
V = 32.1 / 3.107 = 10.33m/s
kinetic energy of the system after collision =
½m1v² + ½m2v²
K.E = ½(m1 + m2)v²
K.E = ½(0.107+3) * 10.33²
K.E = 165.77J
Answer: The person will still have a mass of 90kg on Mars
Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.
In this case the person will have a Weight of 90*9.8 = 882N on Earth.
{ "g" on Earth is 9.8m/s²}
And a Weight of 90*3.3 = 297N on Mars.
{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}
From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.
Answer:
I believe it's the bottom left :)
Explanation:
Answer:
g = 4.7 ×
m/
Explanation:
Given that the mass of the satellite = 700 kg, and 10,000 m above the earth;s surface.
From Newton's second law,
F = mg ............... 1
From Newton's gravitation law,
F =
.................. 2
Where: F is the force, G is the gravitational constant, M is the mass of the first body, m is the mass of the second body, g is the gravitational force and r is the distance between the centers of the two bodies.
Equate 1 and 2 to have,
mg = 
⇒ g = 
But; G = 6.67 ×
N
, M = 700 Kg, r = 10000 m
Thus,
g = 
= 
= 4.669 ×
m/
The force of gravity on the satellite is 4.7 ×
m/
.