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Mandarinka [93]
3 years ago
5

The condition required to work to be done​

Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

work=f(costheta)

Explanation:

work is done when a force acts on a body and displaces it on the direction of force

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So is it the last one?
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Answer:

last what? ion see nothing

Explanation:

hi

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3 years ago
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When sound waves travel through a medium, the particles of the medium move parallel to the wave motion. Which type of wave displ
Karo-lina-s [1.5K]
Longitudinal waves move parallel to the wave motion.
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3 years ago
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A car bounces up and down on its springs at 1.0 Hz with only the driver in the car. Now the driver is joined by four friends. Th
romanna [79]

The new frequency of oscillation when the car bounces on its springs is 0.447 Hz

<h3>Frequency of oscillation of spring</h3>

The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where

  • k = spring constant and
  • m = mass on spring

Now since k is constant, and f ∝ 1/√m.

So, we have f₂/f₁ = √(m₁/m₂) where

  • f₁ = initial frequency of spring = 1.0 Hz,
  • m₁ = mass of driver,
  • f₂ = final frequency of spring and
  • m₂ = mass on spring when driver is joined by 4 friends = 5m₁

So, making f₂ subject of the formula, we have

f₂ = [√(m₁/m₂)]f₁

Substituting the values of the variables into the equation, we have

f₂ = [√(m₁/m₂)]f₁

f₂ = [√(m₁/5m₁)]1.0 Hz

f₂ = [√(1/5)]1.0 Hz

f₂ = 1.0 Hz/√5

f₂ = 1.0 Hz/2.236

f₂ = 0.447 Hz

So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz

Learn more about frequency of oscillation of spring here:

brainly.com/question/15318845

7 0
2 years ago
A baseball has a mass of 0.15kg .What is the net force on the ball if it's acceleration is 40.0m/sw
kvasek [131]

By Newton's second law, we have

F=ma

So, in order to give a 0.15kg body an acceleration of 40m/s^2, you need a force of

F=0.15\cdot 40 = 6N

7 0
4 years ago
A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B f
Misha Larkins [42]

Answer:

A) False

B) False

C) True

D) False

Explanation:

A) False, because when leaving the field, the coil experiences a magnetic force to the right.

B) When the loop is entering the field, the magnetic flux through it will increase. Thus, induced magnetic field will try to decrease the magnetic flux i.e. the induced magnetic field will be opposite to the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is out of the plane of figure. Due to that reason, the current would be counterclockwise. So the statement is false.

C) When the loop is leaving the field, the magnetic flux through the loop will decrease. Thus, induced magnetic field will try to increase the magnetic flux i.e. the inducued magnetic field will be in the same direction as the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is also into the plane of figure. Due to that reason, the current would be clockwise. So the statement is true.

D) False because when entering the field magnetic force will be toward left side

8 0
3 years ago
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