Answer
given,
initial velocity of skateboard = 5.1 m/s
angle above the horizontal = 55°
height of the ramp = 1 m
a) maximum height of projectile
H = 0.889 m
the maximum height of the skateboard above the ground
= 1 + 0.889
= 1.889 m
b) time to reach the height
t = 0.426 s
horizontal distance = u cos θ × t
= 5.1 × cos 55° × 0.426
horizontal distance = 1.25 m
It would be d all of the above
Answer:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:
If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.
Answer:
Rebounce angle is 345°
Rebounce speed is 989.95m/s
Explanation:
Calculate the x component of the velocity of the bullet before impact by using the following relation:
Vbx= Vb Cos thetha
Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°
Substituting
Vbx = Cos15 ×1400 = 1352.30m/s
Calculate the y component using the relation:
Vby = Vo Sin theta
Vby = sin 15° × 1400
Vby = 362.35m/s
The rebounce angle = 360 - incidence angle
Rebounce angle =( 360 - 15)° = 345°
The rebound speed V' = Vby - Vbx
V' = (1352.30 - 362.35)m/s
V' = 989.95 m/s
Answer:
second is the SI unit of time