First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
Weight = (14 kg)(9.8 m/s²)
Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle.
Weight = (137.2 N)(sin 52°)
weight = 108.1 N
To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline.
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
IMA = 3/1
= 3
Therefore the IMA of the ramp is 3
This means the ramp increases the force that is being exerted by 3 times.
Answer:
3600N
Explanation:
Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s
ΣF = ma
we need to find 'a' first, using the definition of 'a' we get equation:
a = (Vf-Vo)/Δt
a = (30m/s)/10s
a = 3 m/s^2
now substitute into top equation
ΣF = ma
Fengine = (1200kg)(3m/s^2)
Fengine = 3600N
Answer:
Height of cliff = S = 20 m (Approx)
Explanation:
Given:
Initial velocity = 8 m/s
Distance s = 16 m
Starting acceleration (a) = 0
Computation:
s = ut + 1/2a(t)²
16 = 8t
t = 2 sec
Height of cliff = S
Gravitational acceleration = 10 m/s
S = 1/2a(t)²
S = 1/2(10)(2)²
Height of cliff = S = 20 m (Approx)
Answer: 117 kPa
Explanation:
For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa
For the liquid at depth 9m, the gauge pressure is equal to= P₂
Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.
So, For finding gauge pressure we have formula P= ρ * g * h
Also gravity also remains same for both liquids
So taking ratio of their respective pressures we have
= 
So 
= 
Or P₂= 39 * 3 = 117 kPa
