Answer:
![R = 20\,N](https://tex.z-dn.net/?f=R%20%3D%2020%5C%2CN)
Explanation:
The free body diagrams of each box are presented in the images attached below. Given the smoothness of the level floor, friction forces can be neglected and due to the contact (3rd Newton's Law) betweeen each other, both boxes experiment the same acceleration:
Box A
![\Sigma F_{x} = F - R = m_{A}\cdot a](https://tex.z-dn.net/?f=%5CSigma%20F_%7Bx%7D%20%3D%20F%20-%20R%20%3D%20m_%7BA%7D%5Ccdot%20a)
![\Sigma F_{y} = N_{A} - m_{A}\cdot g = 0](https://tex.z-dn.net/?f=%5CSigma%20F_%7By%7D%20%3D%20N_%7BA%7D%20-%20m_%7BA%7D%5Ccdot%20g%20%3D%200)
Box B
![\Sigma F_{x} = R = m_{B}\cdot a](https://tex.z-dn.net/?f=%5CSigma%20F_%7Bx%7D%20%3D%20R%20%3D%20m_%7BB%7D%5Ccdot%20a)
![\Sigma F_{y} = N_{B} - m_{B}\cdot g = 0](https://tex.z-dn.net/?f=%5CSigma%20F_%7By%7D%20%3D%20N_%7BB%7D%20-%20m_%7BB%7D%5Ccdot%20g%20%3D%200)
After some algebraic manipulation:
![F = (m_{A} + m_{B})\cdot a](https://tex.z-dn.net/?f=F%20%3D%20%28m_%7BA%7D%20%2B%20m_%7BB%7D%29%5Ccdot%20a)
The acceleration experiment for each box is:
![a = \frac{F}{m_{A}+m_{B}}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF%7D%7Bm_%7BA%7D%2Bm_%7BB%7D%7D)
![a = \frac{32\,N}{3\,kg + 5\,kg}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B32%5C%2CN%7D%7B3%5C%2Ckg%20%2B%205%5C%2Ckg%7D)
![a = 4\,\frac{m}{s^{2}}](https://tex.z-dn.net/?f=a%20%3D%204%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
The contact between the two boxes is:
![R = (5\,kg)\cdot (4\,\frac{m}{s^{2}})](https://tex.z-dn.net/?f=R%20%3D%20%285%5C%2Ckg%29%5Ccdot%20%284%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%29)
![R = 20\,N](https://tex.z-dn.net/?f=R%20%3D%2020%5C%2CN)
False, m<span>ore massive objects exert a stronger gravitational force.</span>
Answer:
f=156Hz
F=467.25N
wavelength =1.6m
Explanation:
we have two frequencies from the question 624hz and 780hz.
let them be f1 and f2
f1=624=nv/2L
f2=780=(n+1)v/2L
![\frac{f1}{f2}=624/780 =\frac{n}{n+1}](https://tex.z-dn.net/?f=%5Cfrac%7Bf1%7D%7Bf2%7D%3D624%2F780%20%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
624n+624=780n
n=4
velocity=2f1l/4
v=624*0.8/2
V=249.6m/s
The fundamental frequency of a wave is the following
f=v/2L
f=249.6/1.6
f=156hz
v=![\sqrt{T/u}](https://tex.z-dn.net/?f=%5Csqrt%7BT%2Fu%7D)
T=v^2u
=249.6^2*7.5*10^-3
T=467.25N
f=156Hz
T=467.25N
T=tension in the string
wavelength is
v=f w
w=249.6/156
w=1.6m
Answer:
f1 = -3.50 m
Explanation:
For a nearsighted person an object at infinity must be made to appear to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.
∴ v = -3.5 m
Using mirror formula,
i/f1 = 1/v + 1/u
Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u = object distance = at infinity(∞) = 1/0
∴ 1/f1 = (1/-3.5) + 1/infinity
Note that, 1/infinity = 1/(1/0) = 0/1 =0.
∴ 1/f1 = 1/(-3.5) + 0
1/f1 = 1/(-3.5)
Solving the equation by finding the inverse of both side of the equation.
∴ f1 = -3.50 m
Therefore a converging lens of focal length f1 = -3.50 m
would be needed by the person to see an object at infinity clearly
Yes I'm pretty sure you can