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Dafna1 [17]
3 years ago
10

Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en

counter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?
Physics
1 answer:
allsm [11]3 years ago
4 0

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

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5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?
GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

Learn more:

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3 0
3 years ago
How are motors and generators different?
saw5 [17]
A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy
4 0
3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos
faust18 [17]

Answer:

The possible thickness of the soap bubble = 1.034\times 10^{-7}\ m.

Explanation:

<u>Given:</u>

  • Refractive index of the soap bubble, \mu=1.33.
  • Wavelength of the light taken, \lambda = 550.0\ nm = 550.0\times 10^{-9}\ m.

Let the thickness of the soap bubble be t.

It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.

For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:

2\mu t=\left ( m+\dfrac 12 \right )\lambda.

where m is the order of constructive interference.

Since the soap bubble is appearing very bright, the order should be 0, as 0^{th} order interference has maximum intensity.

Thus,

2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.

It is the possible thickness of the soap bubble.

6 0
3 years ago
What is the frequency, in hertz, of a sound wave (v = 340 m/s) with a wavelength of 68 m?
hram777 [196]

Answer:

frequency = 5 Hz

Explanation:

F = v/wavelength

F = 340/68 =5Hz

5 0
3 years ago
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