Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
C. Both reach the bottom at the same time.
Explanation:
For a rolling object down an inclined plane , the acceleration is given below
a = g sinθ / (1 + k² / r² )
θ is angle of inclination , k is radius of gyration , r is radius of the cylinder
For cylindrical object
k² / r² = 1/2
acceleration = g sinθ /( 1 + 1/2 )
= 2 g sinθ / 3
Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.
Heat flows irreversibly from hot to cold
Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
Answer:
Water flowing rate= (300000kg/s) = (300000l/s)
Explanation:
First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:
Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s
With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:
Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)
Taking into account that 1kg of water is approximately equal to 1 liter of water.
