A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in
1 answer:
Answer:
96046 Ns.
Explanation:
We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.
40 km/h in the east
V₁ = 40 i
V₂ = 50j
momentum p₁ = mV₁
= 1500 X 40 i
= 60000 i
Momentum p₂ = mV₂
= 1500 X 50j
= 75000 j
Change in momentum
p₂ - p₁
75000j - 60000i
Magnitude of change
=
= 96046 Ns.
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Answer:
Vertical component of velocity is 9.29 m/s
Explanation:
Given that,
Velocity of projection of a projectile, v = 22 m/s
It is fired at an angle of 22°
The horizontal component of velocity is v cosθ
The vertical component of velocity is v sinθ
So, vertical component is given by :
Hence, the vertical component of the velocity is 9.29 m/s
Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>