Answer:
The tension in the rope is 281.60 N.
Explanation:
Given that,
Length = 3.0 m
Weight = 600 N
Distance = 1.0 m
Angle = 60°
Consider half of the ladder,
let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.
....(I)
.....(II)
On taking moment about base
Put the value into the formula
....(III)
We need to calculate the force for ladder
We need to calculate the tension in the rope
From equation (3)
Hence, The tension in the rope is 281.60 N.
Answer:
1.36
Explanation:
= Index of refraction of air = 1
= Index of refraction of plastic = ?
i = angle of incidence in air = 32.0° deg
r = angle of refraction in plastic = 23.0° deg
Using Snell's law
Sini = Sinr
(1) SIn32.0 = Sin23.0
= 1.36
Answer:
1331.84 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement = 490 km
a = Acceleration
g = Acceleration due to gravity = 1.81 m/s² = a
From equation of linear motion
The speed of the material must be 1331.84 m/s in order to reach the height of 490 km
The labeled points which is Letter B in the given Image is the point that the axis of rotation passes through. This problem is an example of rotational dynamics, formerly an object moves in a straight line then the motion is translational but when an object at inactivity lean towards to continue at inactivity and an object in rotation be possible to continue rotating with continuous angular velocity unless bound by a net external torque to act then is rotational. In a rotational motion, the entity is not treated as a constituent part but is treated in translational motion. It points out with the study of torque that outcomes angular accelerations of the object.
Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
