A car traveling in a straight line has a velocity of 6m/s at some instant. After 6.32s its velocity is 13.2m/s . What is the ave
1 answer:
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Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
C. The bug's change in momentum is equal to the car's change in momentum.
Explanation:
As we know by Newton's 2nd law
here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law
so we know that
so we have
so correct answer will be
C. The bug's change in momentum is equal to the car's change in momentum.