Answer:
An object moves with constant velocity .
Explanation:
•2nd law
Answer:
Frequency = 3 Hz
Explanation:
Frequency is a measure of Hertz. Recall that Hertz is the unit expressing cycles/second, where one second is the denominator of the fraction for simplicity. If there are 12 waves every four seconds, and one wave represents one cycle, dividing 12 waves by 4 seconds will give the answer of 3 waves (or cycles) per one second.
The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:


We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:

And the velocity achieved in part 1:

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

Then, to get
, we do:



And we substitute the values:

Total thermal energy is the answer to your question.