A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s east find the vel
1 answer:
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Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
Second minima from center
The distance between the places where the intensity is zero due to the double slit effect
Put the value into the formula
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
The resultant static friction force is equal to 20 N to the left.
Why?
I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:
"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"
So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.
We can calculate the static friction force by using the following formula:
Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.
Calculating we have:
Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)
So,
Since the static friction force is equal to the applied force, the crate does not start to move.
Have a nice day!
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W=
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules