A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. If the norma
l force exerted on the suitcase is 160 n, what is the force f applied to the handle?
2 answers:
<u>Answer:</u>
The force F applied to the handle = 330.03 N
<u>Explanation:</u>
The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have
Horizontal component of force = F cos θ
Vertical component of force = F sin θ
In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.
So, F sin 29 = 160
F = 330.03 N
The force F applied to the handle = 330.03 N
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Answer:
B = 5.59x10⁹ T
Explanation:
The magnetic force (F), on a the alpha particle with charge (q) that is moving at velocity (v) as the cross product of the velocity and magnetic field (B) is:
<u>We have:</u>
F = 1.4x10⁻³ N
v = 2.6x10⁶ m/s
θ = 37.0°
q = 2*p = 2*1.6x10⁻¹⁹ C
Hence, the strength of the magnetic field is:
Therefore, the strength of the magnetic field is 5.59x10⁹ T.
I hope it helps you!