Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.
The distance covered is 1000 m
Explanation:
The rocket is moving by uniformly accelerated motion, so we can find the distance it covers by using the following suvat equation:

where
s is the distance covered
v is the final velocity
t is the time
a is the acceleration
For the rocket in this problem, we have:
v = 445 m/s is the final velocity
is the acceleration
t = 4.50 s is the time
Substituting, we find the distance covered:

Learn more about accelerated motion:
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The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude <em>w</em>), the normal force (mag. <em>n</em>), and the friction force (maximum mag. <em>f</em> ).
In the horizontal direction, we have
<em>n</em> cos(120º) + <em>f</em> cos(30º) = 0
-1/2 <em>n</em> + √3/2 <em>f</em> = 0
<em>n</em> = √3 <em>f</em>
and in the vertical,
<em>n</em> sin(120º) + <em>f</em> sin(30º) + (-<em>w</em>) = 0
<em>n</em> sin(120º) + <em>f</em> sin(30º) = (50 kg) (9.80 m/s²)
√3/2 <em>n</em> + 1/2 <em>f</em> = 490 N
Substitute <em>n</em> = √3 <em>f</em> and solve for <em>f</em> :
√3/2 (√3 <em>f </em>) + 1/2 <em>f</em> = 490 N
2 <em>f</em> = 490 N
<em>f</em> = 245 N
(pointed up the incline)
Answer:
d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.
Explanation:
While moving the bag to the shelf in one shot we can say that the total work done is given as

here we know that
2H = total height raised by the bag
now when we raise the bag to first shelf and then move it to next shelf
then we will have
![W = W_1 + W_2[tex][tex]W = mgH + mgH](https://tex.z-dn.net/?f=W%20%3D%20W_1%20%2B%20W_2%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20mgH%20%2B%20mgH)

so the correct answer will be
d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.
did you mean what activities are performed during the cool down?