LOADS OF POINTS - PLEASE SHOW WORKING AND METHOD USED
2 answers:
The formula your looking for is Vf = Vi + A (T)
Vf = Velocity Final
Vi = Velocity Initial
A = Acceleration
T = Time
Our Initial Velocity is 15m/s. That’s our starting point.
Acceleration is -3m/s. It’s negative because we’re slowing down
We know that the cyclist comes to a complete stop, which means the Final Velocity is 0
In order to solve this with the equation. We need three of the four variables. We just don’t know time. So our setup is -
0 = 15-3(T)
Then you solve
Vf = Velocity Final
Vi = Velocity Initial
A = Acceleration
T = Time
Our Initial Velocity is 15m/s. That’s our starting point.
Acceleration is -3m/s. It’s negative because we’re slowing down
We know that the cyclist comes to a complete stop, which means the Final Velocity is 0
In order to solve this with the equation. We need three of the four variables. We just don’t know time. So our setup is -
0 = 15-3(T)
Then you solve
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Answer:
a)
b)
c)
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,
- radius of the Mars,
- mass of human,
a)
Gravitation force exerted by the Mars on the human body:
where:
= gravitational constant
b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.
c)
When a similar person of the same mass is standing at a distance of 4 meters:
d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m