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:
0.76
Explanation:
we are given:
radius (r) =5.7 m
speed (s) = 1 revolution in 5.5 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
coefficient of friction (Uk) = ?
we can get the minimum coefficient of friction from the equation below
centrifugal force = frictional force
m x r x ω^{2} = Uk x m x g
r x ω^{2} = Uk x g
Uk =
where ω (angular velocity) =
= = 1.14
Uk = = 0.76
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.