Here, you can derive that by numerical method, as follows:
F = m.a
m = F/a
So, here we can see when we decrease one, other increase by same effect; we can say they are "Indirectly Proportional" to each other!
Hope this helps!
Answer:
Explanation:We should know that weight = mass * gravity.
That is weight equals mass times gravity.
Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.
Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².
On Earth, your weight is 70 kg = W
W = mass x 9.8
70 = mass x 9.8
Your mass is 70/ 9.8
i.e approximately 7.14
Weight at the Moon, W' = 7.14 x 1.62
Hence, your weight on the surface of the moon is just 11.56 kg.
Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!
M= Height of image/height of the object
If, M>1, then height of image>height of object.
And therefore, image is larger than the object.
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Explanation:
Initial energy = final energy + work done by friction
PE = PE + KE + W
mgH = mgh + 1/2 mv² + W
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v² + 25000
v = 22.1 m/s
Without friction:
PE = PE + KE
mgH = mgh + 1/2 mv²
(800)(9.8)(30) = (800)(9.8)(2) + 1/2 (800) v²
v = 23.4 m/s
