Answer:
a = - 3.75 m/s²
negative sign indicates deceleration here.
Explanation:
In order to find the constant deceleration of the car, as it stops, we will use the 3rd equation of motion. The 3rd equation of motion is as follows:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration of the car = ?
Vf = Final Velocity = 0 m/s (Since, the car finally stops)
Vi = Initial Velocity = 30 m/s
s = distance covered by the car = 120 m
Therefore,
a = [(0 m/s)² - (30 m/s)²]/(2)(120 m)
<u>a = - 3.75 m/s²</u>
<u>negative sign indicates deceleration here.</u>
Answer:
We could get the time taken by the ball to return back to earth, using the formula:
s = u t + ½ a t², where
s = displacement of the body moving with initial velocity u, acceleration 'a' in time t.
In the present case s=0 (as the ball returns back to starting time)
u= 30 m/s; a = -10 m/s² ( negative sign as a is in opposite direction to u); t=?
0 = 30 t - ½ ×10 ×t²; ==> 5 t = 30, t= 6 second.
So ball will return back after 6 second after being thrown up.
Explanation:
I looked it up
Hope this helps
Answer:
4360 Kgm/s
Explanation:
Applying,
Ft = M-M'................. Equation 1
Where F = force, t = time, M = Final momentum, M' = Initial momentum.
make M the subject of the equation
M = Ft+M'............ Equation 2
From the question,
Given: F = 4000 N, t = 0.9 seconds, M' = 400 kg-m/s
Substitute these values into equation 2
M = 4000+(0.9×400)
M = 4000+360
M = 4360 kgm/s
Hence the final momentum is 4360 kgm/s
