Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
Answer: option D) 42.4 N
The weight of the frame is balanced by the vertical component of tension.
W = T sin θ + T sin θ = 2 T sin θ
The tension in each cable is T = 30 N
Angle made by the cables with the horizontal, θ = 45°
⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N
Hence, the weight of the frame is 42.4 N. Correct option is D.
Answer:
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Answer:
F = 69.5 [N]
Explanation:
We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

where:
N = normal force [N]
miu = friction coefficient
f = friction force = 22 [N]
Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

where:
F = force exerted [N]
f = friction force [N]
m = mass = 95 [kg]
a = acceleration = 0.5 [m/s²]
Now replacing:
![F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]](https://tex.z-dn.net/?f=F%20-%2022%20%3D%2095%2A0.5%5C%5CF%20%3D%2047.5%20%2B%2022%5C%5CF%20%3D%2069.5%20%5BN%5D)
Answer:
50.4 N
Explanation:
Q1 = Q
Q2 = 4 Q
Distance = d
The force is given by

.... (1)
Now,
Q3 = 2 Q
Q4 = 7 Q
distance = d/3

.... (2)
Divide equation (2) by equation (1), we get
F' / 1.60 = 126 / 4
F' = 50.4 N
Thus, the force is 50.4 N.