We Know, F = m*a
Here, F = 34 N
m = 213 Kg
Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²
So, your final answer & the acceleration of the object would be 0.159 m/s²
Hope this helps!
Answer:
See Explanation
Explanation:
a) We know that;
v = λf
Where;
λ = wavelength of the wave
f = frequency of the wave
v = velocity of the wave
So;
T = 2 * 2.10 s = 4.2 s
Hence f = 1/4.2 s
f = 0.24 Hz
The wavelength = 6.5 m
Hence;
v = 6.5 m * 0.24 Hz
v = 1.56 m/s
b)The amplitude of the wave is;
A = 0.600 m/2 = 0.300 m
c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same
Where d = 0.30 m
A = 0.30 m/2 = 0.15 m
The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
Explanation:
The expression is :
A =[LT], B=[L²T⁻¹], C=[LT²]
Using dimensional of A, B and C in above formula. So,
![A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}](https://tex.z-dn.net/?f=A%3DB%5EnC%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3D%5BL%5E2T%5E%7B-1%7D%5D%5En%5BLT%5E2%7D%5D%5Em%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%7DT%5E%7B-n%7DL%5EmT%5E%7B2m%7D%5C%5C%5C%5C%5C%20%5BLT%5D%3DL%5E%7B2n%2Bm%7DT%5E%7B2m-n%7D)
Comparing the powers both sides,
2n+m=1 ...(1)
2m-n=1 ...(2)
Now, solving equation (1) and (2) we get :
Hence, the correct option is (E).
Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
