Answer: D.) 39,200 J
Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:
P.E.= 39,200 Joules
Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
Force is equal to mass multiplied by acceleration, therefore
F=ma
m=2569.6 kg
a=4.65m/s^2
therefore F=2569.6*4.65=11948.6 (correct to 1 d.p.)
Answer:
Explanation:
velocity of first projectile after 3 s
v = u - gt
v = 49.4 - 9.8 x 3
= 20 m /s
Velocity of second projectile after 3 s after being dropped from rest
v = u + gt
= 0 + 9.8 x 3
= 29.4 m /s
They will be moving in opposite direction at the time of meeting , so their relative velocity
= 20 + 29.4 = 49.4 m /s
From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .
Answer:5
Explanation:
Decimal :2
Significant notation :2.4632× 10^2
