A paper airplane with mass 0.1 kg is flying 1.5 m above the ground with a speed of 2 m/s. what is the total mechanical energy of
1 answer:
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Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Answer:
a) KE = 888.26J
b) N = 294.5 turns
Explanation:
For the kinetic energy:
The inertia is:
So, the kinetic energy will be:
Now, friction force is:
Ff = μ*N = 0.80*5N = 4N
The energy balance would be:
Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.
Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.
Replacing these values into the energy balance:
0-888.26=-4*N*2*π*0.12
-888.26=-0.96*π*N
N=294.5 turns
Answer:51.44 units
Explanation:
Given
x component of vector is
y component of vector is
so position vector is
Magnitude of vector is
|r|=51.44 units
Direction
vector is in 2nd quadrant thus