Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed,
rev/sec
rad/sec
= 5.89 rad/sec
Therefore, force required,

= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
Answer:
Weight of the dog on surface of earth is 140.14 Newton.
Given:
mass of the dog = 14.3 kg
To find:
Weight of the dog = ?
Formula used:
Weight of the dog is given by,
W = mg
Where, W = weight of the dog
m = mass of the dog
g = acceleration due to gravity
Solution:
Weight of the dog is given by,
W = mg
Where, W = weight of the dog
m = mass of the dog = 14.3 kg
g = acceleration due to gravity
W = 14.3 × 9.8
W = 140.14 Newton
Weight of the dog on surface of earth is 140.14 Newton.
Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.
Answer:
<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

- v is image distance
- u is object distance, u is 10 cm
- f is focal length, f is 5 cm

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>
• Let's derive this formula from the lens formula:

» Multiply throughout by fv

• But we know that, v/u is M

- v is image distance, v is 10 cm
- f is focal length, f is 5 cm
- M is magnification.

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>
- Image is magnified
- Image is erect or upright
- Image is inverted
- Image distance is identical to object distance.
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 

where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector


(b) The length of the string is given by:


(c) Now, we first find the frequency of the wave:



Now,
Speed of the wave is given by:

