If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
Answer:
P = 4000 [Pa]
Explanation:
Pressure is defined as the relationship between Force and the area where the body rests.
The support area is equal to:
![A=50*20=1000[cm^{2} ]](https://tex.z-dn.net/?f=A%3D50%2A20%3D1000%5Bcm%5E%7B2%7D%20%5D)
But we must convert from square centimeters to square meters.
![1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]](https://tex.z-dn.net/?f=1000%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%7B100%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%3D0.1%5Bm%5E%7B2%7D%20%5D)
And the pressure is:
![P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5CP%3D400%2F0.1%5C%5CP%3D4000%5BN%2Fm%5E%7B2%7D%20%5Dor%204000%5BPa%5D)
Answer:
magnitude of force on charge 2Q = 
Direction of force on charge = 61 ⁰
Explanation:
The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force
║F║ =
= after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW
magnitude of force acting on 2Q = 
The direction of the force on charge 2Q is calculated as
tan ∅ =
= 1.8284
therefore ∅ =
1.8284
= 61⁰