Answer:
19.95 J
Explanation:
The center of mass of the ladder is initially at a height of:
The center of mass of the ladder ends at a height of:
=L/2
So, the work done is equal to the change in potential energy which is:
W = PE =
now
therefore
W = [mgL/2]×[1 - sin(theta)]
W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]
solving this we get
W = 19.95 J
Answer:
Induced current I = 0.44 A
Explanation:
given data
no of turn N = 18
radius = 27.0 cm
resistance of the coil = 4.50 Ω
time = 2.90 s
magnetic field = 1.90 T to 0.500 T
solution
we get Induced EMF that is
E = - dφ ÷ dt ..............1
E = - N × A × (B2 - B1) ÷ Δt
E = N × A × (B1 - B2) ÷ Δt
area of cross section of the coil A = π × r²
area of cross section = 3.14 × (0.27)²
area of cross section = 0.228 m²
Initial magnetic field B1 = 1.9 T
and
Final magnetic field B2 = 0.5 T
time Δt = 3.7 s
put here value and we get
E =
E = 1.98 V
and
Induced current I is express as
Induced current I = E ÷ R ................2
Induced current I =
Induced current I = 0.44 A
1) 12.5 N east
There are two forces acting on the box along the horizontal direction:
- The applied force of 15 N east, we indicate it with F
- The force of friction of 2.5 N west, we indicate it with
Taking east as positive direction, we can write the two forces has
Therefore, the net force on the box will be:
And the positive sign means the direction is east.
2)
We can solve this part by using Newton's second law:
where
is the net force on the box
m is its mass
a is the acceleration
For the box in this problem,
(east)
m = 5.0 kg
Solving for a, we find the acceleration:
And the direction is the same as the net force (east)