Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
Weight is a force so by F = ma, we have
weight = 10 x 1.6
= 16 N
A model of what Shoulder pain looks like...
Answer:
attribute is defined as a quality or characteristic of a person, place, or thing
Explanation:
Answer:
36.22 mA
Explanation:
i1 = I , i2 = I, d = 8.2 cm = 0.082 m
Force per unit length = 3.2 nN/m = 3.2 x 10^-9 N/m
μo = 4 π × 10^-7 Tm/A
The formula for the force per unit length between the two wires is given by
F = μo / 4π x (2 i1 x i2) / d
3.2 x 10^-9 = 10^-7 x 2 x I^2 / 0.082
I = 0.0362 A = 36.22 mA