<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16
Traveling against the current:
(18 - c)*t = 8
Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16
Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16
(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0
Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c
So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3
(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3
And it's verified.</span>
Answer:
See below
Explanation:
Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N
no work is done by this force
Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N
work of friction = 7.55 * 2 m = 15.1 j
Force Downplane = mg sin 53 = 62.61 N
work = 62.61 * 2 = 125.22 j
Net Force downplane = force downplane - force friction = 55.06 N
net Work = force * distance = 55.06 N * 2 M = 110.12 j
Answer:
static coefficient = 0,203 & kinetic coefficient = 0,14
Explanation:
There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.
In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.
In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.
