Solution: From the given question, we shall find the vector quantity among the
(A) Time , (B) Velocity, (C) Distance , (D) Speed
Concept: <u>Vector Quantity: </u>All those physical quantities which have magnitude as well as specific directions, are called Vector Quantities.
Here, Time, Distance and Speed have only magnitude but have no directions so they will be scalar quantities.
Now, <u>Velocity:</u> It is defined as the change in displacement per unit time. Since the change in the displacement will be in particular direction only. Hence, velocity will be the vector quantity.
Hence, the option (B) Velocity will be the correct option.
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at 
Solving for the angle, we get
or
![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)
The answer:
<span>When the elevator accelerates upward at a rate of 3.6 m/s², the value of the acceleration becomes
</span>A=g+3.6=13.4 m/s²
and by using the newton's law, F=mass x A, we have
T1= (24 + 90 )x 13.4= 1527.6 N, where T1 is the <span>Tension in upper rope
</span> and
T2= ( 90 )x 13.4= 1206N, where T2 is the Tension in lower rope
When the elevator accelerates downward at a rate of 3.6 m/s², the value of the acceleration becomes
A=9.8 - 3.6 = 6.2 m/s²
T1= (24 + 90 )x 6.2= 706.8 N, where T1 is the Tension in upper rope
and
T2= ( 90 )x 6.2= 558N, where T2 is the Tension in lower rope
