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 .
Answer:
The horizontal range will be 
Explanation:
We have given initial speed of the shell u = 
Angle of projection = 51°
Acceleration due to gravity 
We have to find maximum range
Horizontal range in projectile motion is given by
So the horizontal range will be
Answer: 2.7 m/s
Explanation:
Given the following :
Period (T) = 8.2 seconds
Radius = 3.5 m
The tangential speed is given as:
V = Radius × ω
ω = angular speed = (2 × pi) / T
ω = (2 × 22/7) / 8.2
ω = 6.2857142 / 8.2
ω = 0.7665505
Therefore, tangential speed (V) equals;
r × ω
3.5 × 0.7665505 = 2.6829268 m/s
2.7 m/s
Answer:
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Explanation:
