The troposphere is the layer of the atmosphere which contacts the surface of the Earth. This layer is <span>0 to 12 km (0 to 7 miles).
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Answer:
5.05 m/s
Explanation:
The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:
Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m
Answer:
4.32
Explanation:
The centripetal acceleration of any object is given as
A(cr) = v²/r, where
A(c) = the centripetal acceleration
v = the linear acceleration
r = the given radius, 1.9 m
Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that
9.8 = v²/1.9
v² = 1.9 * 9.8
v² = 18.62
v = √18.62
v = 4.32 m/s
This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s
<u>Answer:</u>
If a car weighs 6310 N on earth, its mass = 643.22 kg.
<u>Explanation:</u>
Weight of a body is given by the product of it's mass and acceleration due to gravity on that planet.
So weight of body in earth = mass of body* acceleration due to gravity on earth.
Acceleration due to gravity in earth =
Here it is given weight of car = 6310 N
So we have, 6310 = mass of car* 9.81
Mass of car = 6310/9.81 = 643.22 kg.
So If a car weighs 6310 N on earth, its mass = 643.22 kg.