Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.
Explanation:
First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.
From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.
Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.
Answer: A
Explanation:
Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.
Answer:
12 N/cm²
Explanation:
From the question given above, the following data were obtained:
Weight (W) of block = 240 N
Area (A) = 20 cm²
Pressure (P) =?
Next, we shall determine the force exerted by the block. This can be obtained as follow:
Weight (W) of block = 240 N
Force (F) =.?
Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,
Force (F) = Weight (W) of block = 240 N
Force (F) = 240 N
Finally, we shall determine the pressure on the floor as follow:
Force (F) = 240 N
Area (A) = 20 cm²
Pressure (P) =?
P = F/A
P = 240 / 20
P = 12 N/cm²
Therefore, the pressure on the floor is 12 N/cm².
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :
v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
Answer:
Answered
Explanation:
v= 1 m/s
A= 1 m^2
m= 100 kg
y= 1 mm
μ = ?
ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2
forces acting on the block are
F_s ← ↓ →F_f
mg
N= mg
F_s= shear force = ζAv/y F_f= friction force = μN
now in x- direction F_s= F_f
ζAv/y = μN
0.3075×1×1×1/1×10^{-3} = μ×100
⇒μ=0.313 (coefficient of sliding friction for the block)
Now, as the velocity is increased shear force also increases and due to this frictional force also increases.
Now, to compensate this frictional force friction coefficient must increase
as v∝μ
