the radiogenic heat produced by the radioactive decay of isotopes in the mantle and crust, and the primordial heat left over from the formation of the Earth.
Answer:
where L is the length of the ramp
Explanation:
Let L (m) be the length of the ramp, and g = 9.81 m/s2 be the gravitational acceleration acting downward. This g vector can be split into 2 components: parallel and perpendicular to the ramp.
The parallel component would have a magnitude of

We can use the following equation of motion to find out the final velocity of the book after sliding L m:

where v m/s is the final velocity,
= 0m/s is the initial velocity when it starts from rest, a = 2.87 m/s2 is the acceleration, and
is the distance traveled:


Answer:
The back end of the vessel will pass the pier at 4.83 m/s
Explanation:
This is purely a kinetics question (assuming we're ignoring drag and other forces) so the weight of the boat doesn't matter. We're given:
Δd = 315.5 m
vi = 2.10 m/s
a = 0.03 m/s^2
vf = ?
The kinetics equation that incorporates all these variables is:
vf^2 = vi^2 + 2aΔd
vf = √(2.1^2 + 2(0.03)(315.5))
vf = 4.83 m/s
Answer:
56 m/s
Explanation:
The time we are considering is
t = 15 s
The vertical velocity of the projectile is given by

where
is the initial vertical velocity
is the acceleration due to gravity
Substituting t=15 s, we find the vertical velocity of the projectile at that time:

where the negative sign means the direction is now downward.
The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:

So, the magnitude of the velocity at the moment of impact is
