I’m tryna pass this class so pleaseee
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Answer:
W = 1.432 KJ
Explanation:
given,
mass = 22.2 Kg
angle of the rope = 27.5°
distance on the ground = 24 m
kinetic friction= μ = 0.32
acceleration due to gravity, g = 9.8 m/s²
Work done = ?
W = F d cosθ
a = 0 because it is moving with constant speed
equating all the forces acting in x direction
F cosθ = F friction = μN
equating all the forces acting in y direction
F sinθ + N -mg =0
now,
N = mg - F sinθ
putting value of N
F cosθ = μ mg -μ F sinθ
F (cosθ + μsinθ ) = μ mg
F =67.28 N
now,
W=F d cosθ
W =67.28 x 24 x cos(27.5)
W =1432.27 J
W = 1.432 KJ
It is given that an<span> airplane is flying through a thundercloud at a height of 2000 m.
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Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:
</span>
</span><span>
Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:
</span>