Answer:
The two forces acting on the object are weight due to gravity pulling the object towards earth, and drag resisting this motion. When the object is first released, drag is small as velocity is low, so the resultant force is down. This means the object accelerates towards earth.
Answer:
14 m/s²
Explanation:
Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.
(4N + 10 x 1kg)=(1kg)a
14/1=14, so the acceleration is 14 m/s²
Answer:
3/5 v
Explanation:
The computation of speed will the alpha particle have after the collision is shown below:-
In a perfectly elastic the kinetic energy and collision the momentum are considered.
The velocity of the particles defines the below equation:
As we know that
Here, we consider A is the alpha particle and B is the proton and now by the above values we can solve the equation which is below:-
Therefore the correct answer is 
The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
<h3>Acceleration of the box</h3>
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
<h3>Time of motion of the box</h3>
The time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
<h3>Average power supplied by the friction</h3>
P = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
Learn more about average power here: brainly.com/question/19415290
#SPJ1
Answer:
<h2>a)
Acceleration is 3.09 m/s²</h2><h2>
b) Distance traveled is 45.05 m</h2><h2>c)
Time taken to travel 250 m is 12.72 s</h2>
Explanation:
a) We have equation of motion v = u + at
Initial velocity, u = 0 km/hr = 0 m/s
Final velocity, v = 60 km/hr = 16.67 m/s
Time, t = 5.4 s
Substituting
v = u + at
16.67 = 0 + a x 5.4
a = 3.09 m/s²
Acceleration is 3.09 m/s²
b) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 3.09 m/s²
Time, t = 5.4 s
Substituting
s = ut + 0.5 at²
s = 0 x 5.4 + 0.5 x 3.09 x 5.4²
s = 45.05 m
Distance traveled is 45.05 m
c) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 3.09 m/s²
Displacement, s = 0.25 km = 250 m
Substituting
s = ut + 0.5 at²
250 = 0 x t + 0.5 x 3.09 xt²
t = 12.72 s
Time taken to travel 250 m is 12.72 s
