Answer:
θ =28.07⁰
Explanation:
when van moves around an unbanked curve, horizontal component of normal force is equal to centripetal force
---(1)
There is no motion in vertical direction, so vertical component of normal force is balanced with weight of block i.e.
----(2)
divide (1) and (2)
v = 28 m/s
r = 150 m
g = 9.8m/
θ = 28.07⁰
Answer: 1666J
Explanation:
Given that,
Mass of box (m) = 10kg
Total distance covered by box (h)
= (5m + 12m)
= 17m
work done on the box = ?
Work is done when force is applied on an object over a distance. Hence, the magnitude of work done on the box depends on its mass (m), distance covered (h), and acceleration due to gravity (g)
(g has a value of 9.8m/s²
i.e Work = mgh
Work = 10kg x 9.8m/s² x 17m
Work = 1666J
Thus, 1666 joules of work was done by the woman on the box.
Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
Answer:
The speed of the stone just before it hits the ground is 18.54 m/s
Explanation:
Given that,
Initial speed of the stone, u = 8 m/s
The stone is thrown downward from a height of 14 m
We need to find the speed of the stone just before it hits the ground. It can be calculated using third equation of motion as :
v is the speed of the stone just before it hits the ground
v = 18.54 m/s
So, the speed of the stone just before it hits the ground is 18.54 m/s. Hence, this is the required solution.
