Let north as positive
Fnet=10n-5n
=5n north
Answer:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force

Put the value into the formula


So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force

Put the value in to the formula


Hence, The frictional force acting on the block is 14.8 N.
Answer:
<em>The y component of his displacement is 11.22 meters</em>
Explanation:
<u>Components of the displacement</u>
The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

The vertical component is calculated by:

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:


The y component of his displacement is 11.22 meters
Answer:
105 m/s
Explanation:
Given that the speed of train A,
= 45 m/s from west to east.
Speed of train B,
= 60 m/s from east to west.
Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.
So, the speed of train A from east to west= - 45 m/s
The speed of train B w.r.t train A
m/s
Hence, the speed of train B w.r.t train A is 105 m/s from east to west.