Answer:
-1.03 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².
Mathematically, acceleration is expressed as
a = (v-u)/t ........................ Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time.
Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.
Substituting into equation 2
a = (7.20-13.60)/6.2
a = -6.4/6.2
a = -1.03 m/s²
Note: a is negative because, the hockey puck is decelerating.
Hence the average acceleration = -1.03 m/s²
Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.
We have that the sea level pressure for Leh area is 1150mb mathematically given as
Ps= 1150 mb
<h3>
Sea level pressure</h3>
Question Parameters:
Ladakh is 800 mb.
<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m
increase in height with respect to sea level corresponds to 10 mb pressure,
Generally, for 3500m the pressure change will be 350 mb.
Therefore, here for the sea level <em>pressure</em> we need to add,
Ps=800+350
Ps= 1150 mb
For more information on Pressure visit
brainly.com/question/25688500
The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N, 
= 25 N, a = -0.9 
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.
= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
