Answer:
Acceleration=4m/s²
Force applied =619.8N
Explanation:
Using equation of motion
V=u+at we have: u=o, v=50m/s
50= 0 + a×0.0121
a = 50/0.0121
a= 4m/s²
Neglecting resistance forces
F= ma, where a = v-u/t
F=m×(v-u)/t
F= 0.150 ×(50-0)/0.0121
F=7.5/0.0121
F= 619.8N
Answer:
yes, should be
Explanation:
This is a hard yes or no question becuase the amplitudes are the same height but in different beating orders.
Answer:
The answer is 6.40 meters.
Explanation:
The speed v = √(2gh)
v = √( 2×9.8×6.4) = 11.2 m/s
After, finding the time it takes to hit the ground from a height of 1.6 meters.
time = √(2H÷g)
time = √(2×1.6÷9.8)
time = 0.5714 seconds.
Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
