Answer:
Point motion will eventually stops due to action of g exactly perpendicular...
Explanation:
If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.
Conduction, Convection, Radiation.
Hope this helps!
-Payshence
Answer:
W = 20 J
Explanation:
Given that,
The mass of a loaded sack, m = 1000 g = 1 kg
It falls down from the floor of a lorry 200 cm high, h = 2 m
We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :
W = mgh
Substitute all the values,
W = 1 × 10 × 2
= 20 J
Hence, the required work done by gravity is equal to 20 J.
Answer:
θ = 6.3 *10³ revolutions
Explanation:
Angular acceleration of the drill
We apply the equations of circular motion uniformly accelerated
ωf= ω₀ + α*t Formula (1)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (s)
Data
ω₀ = 0
ωf = 350000 rpm = 350000 rev/min
1 rev = 2π rad
1 min= 60 s
ωf = 350000 rev/min =350000*(2π rad/60 s)
ωf = 36651.9 rad/s
t = 2.2 s
We replace data in the formula (2) :
ωf= ω₀ + α*t
36651.9 = 0 + α* (2.2)
α = 36651.9 / (2.2)
α = 17000 rad/s²
Revolutions made by the drill
We apply the equations of circular motion uniformly accelerated
ωf²= ω₀ ²+ 2α*θ Formula (2)
Where:
θ : Angle that the body has rotated in a given time interval (rad)
We replace data in the formula (2):
(ωf)²= ω₀²+ 2α*θ
(36651.9)²= (0)²+ 2( 17000 )*θ
θ = (36651.9)²/ (34000 )
θ = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)
θ = 6288.31 revolutions
θ = 6.3 *10³ revolutions