Answer:
Explanation:
We shall apply conservation of mechanical energy .
initial kinetic energy = 1/2 m v²
= .5 x m x 12 x 12
= 72 m
This energy will be spent to store potential energy . if h be the height attained
potential energy = mgh , h is vertical height attined by block
= mg l sin20 where l is length up the inclined plane
for conservation of mechanical energy
initial kinetic energy = potential energy
72 m = mg l sin20
l = 72 / g sin20
= 21.5 m
deceleration on inclined plane = g sin20
= 3.35 m /s²
v = u - at
t = v - u / a
= (12 - 0) / 3.35
= 3.58 s
it will take the same time to come back . total time taken to reach original point = 2 x 3.58
= 7.16 s
<h3>Answer: 500 Kilometers</h3>
Explanation:
The formula for finding distance is: speed × time
We must multiply:
200Km/h × 2.5 hours
This gives you 500 Km.
<u>Remember: the formula to work out the distance - speed x time.</u>
If the train speed is 200 km/h, it travels 200 km/h.
Now, we must multiply 200 by 2.5:
200 x 2.5 = 500
Our answer is 500 Kilometers.
If the force is 3,000 N and the acceleration is 2 m/s/s. The mass of the car is 1,500 grams (or whatever unit of mass or weight is used in the answer- pounds, kilograms e.t.c.).
Answer:
(a) a = - 201.8 m/s²
(b) s = 197.77 m
Explanation:
(a)
The acceleration can be found by using 1st equation of motion:
Vf = Vi + at
a = (Vf - Vi)/t
where,
a = acceleration = ?
Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)
Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s
t = time = 1.4 s
Therefore,
a = (0 m/s - 282.53 m/s)/1.4 s
<u>a = - 201.8 m/s²</u>
<u></u>
(b)
For the distance traveled, we can use 2nd equation of motion:
s = Vi t + (0.5)at²
where,
s = distance traveled = ?
Therefore,
s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²
s = 395.54 m - 197.77 m
<u>s = 197.77 m</u>
