Answer:
1.503 J
Explanation:
Work done in stretching a spring = 1/2ke²
W = 1/2ke²........................... Equation 1
Where W = work done, k = spring constant, e = extension.
Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.
Substitute into equation 1
W = 1/2(26)(0.34²)
W = 13(0.1156)
W = 1.503 J.
Hence the work done to stretch it an additional 0.12 m = 1.503 J
Answer:
The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.
Explanation:
Given:
Mass of the piano
= 190 kg
Inclined angle
= 18 degree
Considering gravity,
= 9.8
And
Using,
and 
<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>
We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.
a.
When the man pushes it parallel to the incline.
Balancing the forces as
.
⇒ 
⇒ 
⇒ Here it is negative as the force is acting downward.
⇒ Plugging the values of mass
and angle
.
⇒ 
⇒
N
b.
When the force is parallel to the floor.
⇒ 
⇒ 
⇒ Plugging the values.
⇒ 
⇒
N
So,
The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.
Answer:C
Explanation:
Partially submerged block along with vessel is accelerated upwards .
Initially the block weight is supported by buoyant force such that it is in equilibrium.
when the system start accelerating upwards then the effective gravity will be
g+a where a is the acceleration of the system.
so only net gravity is increased so block will not ascend or descend.
Mathematically


where
density of liquid
V=volume of object inside the water
Sure ,Let's find angle between forces
- Vectors be A and B and resultant be R








Energy/power is not gained or lost going through a (ideal) transformer.
So the transformer in this problem really doesn't matter. If the lamp is using energy at the rate of 60 watts, then the whole contraption is getting 60 watts of power from the wall outlet.
Power = (voltage) x (current)
60 watts = (120 v) x (current)
Current = (60 watts) / (120 v)
<em>Current = 0.5 Ampere</em>