Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Answer:
where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please
A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts moving the ball moves toward the back of the train. This happened due to inertia
An object at rest remains at rest, or if in motion, remains in motion unless a net external force acts on it .
When a train starts moving forward, the ball placed on the floor tends to fall backward is an example of inertia of rest. Due to the reason that the lower part of the ball is in contact with the surface and rest of the part is not . As the train starts moving, its lower part gets the motion as the floor starts moving but the upper part will remain as it is as it is not in contact with the floor , hence do not attain any motion due to the inertia of rest simultaneously i.e. it tends to remain at the same place.
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Answer:
0.438kg/ms-¹
Explanation:
Momentum, denoted by p, can be calculated by using the formula;
p = mv
Where;
m = mass (kg)
v = velocity (m/s)
Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹
Momentum (p) of crawling baby = 7.29 kg kg × 0.234 m/s = 1.706kg/ms-¹
Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:
{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
