thats how it works and thanks for points
Answer:
1.38*10^18 kg
Explanation:
According to the Newton's law of universal gravitation:

where:
G= Gravitational constant (6.674×10−11 N · (m/kg)2)
ma= mass of the astronaut
mp= mass of the planet

so:

Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)
Answer:
250Nm
Explanation:
Given parameters:
Length of the long pry bar = 1m
Force acting on it = 250N
Angle = 90°
Unknown:
Amount of torque applied = ?
Solution:
Torque is the turning force on a body that causes the rotation of the body.
The formula is given as:
Torque = Force x r Sin Ф
r is the distance
So;
Torque = 250 x 1 x sin 90 = 250Nm
Answer:
I'm pretty sure its B and C
Explanation:
B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon
C bc mass is the same - you can't change it