A large truck collides with a small car. True or False: The truck exerted a greater magnitude force on the car than the car exer
1 answer:
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Consider:
The mass of the planet is M.
The mass of the orbiting body is m.
The radius of the orbital in which the body is revolving around the planet is r.
The speed of the orbiting body is v.
Solution:
The gravitational force acting on the orbiting body due to the planet is,
where G is the gravitational constant,
This gravitational force acting on the orbiting body helps the body to move in a circular motion around the planet.
The centripetal force acting on the orbiting body is,
As the centripetal force and the gravitational force acting on the orbiting body are the same, thus,
From the formula, the speed of the orbiting body is inversely proportional to the orbital radius.
Thus, with the decrease in the orbital radius of the orbiting body, the velocity of the body increases.
Hence, the relation between the velocity and orbital radius is consistent with the observation in step 8.
Answer:
Explanation:
For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:
v₀ = 48 ft/s
a = -32 ft/s/s
v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:
v² = v₀² + 2aΔx and filling in:
Δx and
0 = 2300 - 64Δx and
-2300 = -64Δs so
Δx = 36 feet.
Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:
v = v₀ + at and filling in:
0 = 48 - 32t and
-48 = -32t so
t = 1.5 sec. That's part a. Onto part b:
The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:
Δx = v₀t + and filling in:
and
and
0 = 16t(3 - t) so
t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:
We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:
and get everything on one side and factor it again:
and we find that
1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.
Answer:
the final speed of the stool is 3.6 rad/s
Explanation:
As we know that there is no external torque on the system
So we can use concept of angular momentum conservation
So we will have
now we will have
So the final speed of the stool is 3.6 rad/s