Answer:
Explanation:
radius of cylinder r₁ = .02 m
radius of safety valve r₂ = .0075 m
force exerted by spring on safety valve = 950 x .0085 = 8.075 N .
Force required on piston of cylinder = F
Applying Pascal's law
8.075 / 3.14 x .0075² = F / 3.14 x .02²
F = 8.075 x .02² / .0075²
= 8.075 x 7.111
= 57.42 N .
Three units deformations observed in these bands.
<h3>What forces do a rubber band encounter?</h3>
Elastic force is the force that permits some materials to regain their former shape after being stretched or crushed. Thus, a stretched rubber band is subject to an elastic force.
The rubber band experiment uses a straightforward rubber band to show entropic force and a refrigeration cycle. The rubber band experiment involves stretching and then releasing a rubber band while measuring its temperature.
Always acting in the opposite direction of motion is friction. This indicates that if friction is there, it cancels out some of the force driving the motion (if the object is being accelerated). This results in a decreased acceleration and a smaller net force.
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Answer:
positive reinforcer; negative reinforcer
Explanation:
positive reinforcer; negative reinforcer
Receiving delicious food can be seen as a positive reinforcer while escaping electric shock can be seen as a negative reinforcer.
The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz
Given Conditions:
mass of string; m = 0.0133 kg
Force on the string; F = 8.89 N
Length of string; L = 1.97 m
1. To find the wavelength at the fourth normal node.
At the fourth harmonic, there will be 2 nodes.
Thus, the wavelength will be;
λ = L/2
λ = 1.97/2
λ = 0.985 m
2. To find the velocity of the wave from the formula;
v = √(F/(m/L)
Plugging in the relevant values gives;
v = √(8.89/(0.0133/1.97)
v = 36.2876 m/s
Now, formula for frequency here is;
f = v/λ
f = 36.2876/0.985
f = 36.84 Hz
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Answer:

Explanation:
Given,
For the first rocket,
- Initial velocity of the first rocket A =

- Acceleration of the first rocket =

For the second rocket,
- Initial velocity of the second rocket B =

- Displacement of both the rockets A and B = s = 0 m
Fro the first rocket,
Let 't' be the time taken by the first rocket A for whole the displacement

Let
be the acceleration of the second rocket B for the same time interval
from the kinematics,


Hence the acceleration of the second rocket B is -33.65\ m/s^2.