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Lana71 [14]
3 years ago
11

Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric

tion inclined plane, and one swings in a circular arc on the end of a string. All three objects end at the same height.
In which case does the object have the biggest total work done on it by all forces during its motion?
A. Free Fall
B. Incline
C. String
D. Same
Physics
1 answer:
erik [133]3 years ago
4 0

Answer:

D. Same

Explanation:

Because only gravity is doing the work on the objects, and gravity is constant for all the objects

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A ____ shows all of the forces acting on an object.
il63 [147K]

Answer:

Free body diagram

Explanation:

A free body diagram shows all the forces acting on a body. Since force is a vector quantity, the magnitude and direction of the forces are shown in the free-body diagram.

Force is the push or pull on a body.

It can be via contact or without contact. Such non-contact forces acts via a force field.

In physics, this free body diagram is used extensively.

7 0
3 years ago
Hannah is very good at picking up differences in other people's mood and behavior. She has been rewarded for her ability to work
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It's C. Most likely
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A 1.5m wire carries a 6 A current when a potential difference of 61 V is applied. What is the resistance of the wire?
Alisiya [41]

Resistance = (voltage) / (current)

For this piece of wire . . .

Resistance = (61 volts) / (6 Amperes)

Resistance = (61/6) (V/A)

<em>Resistance = (10 and 1/6) ohms</em>

Since you know the voltage and current, the length doesn't matter.

7 0
3 years ago
draw a labelled ray diagram to locate the image of an object formed by a convex lens of focal length 20cm when the object is pla
azamat
<h3><u>Answer</u> :</h3>

<u>Focal length of convex lens is taken positive.</u>

:\implies\sf\:\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

:\implies\sf\:\dfrac{1}{v}-\dfrac{1}{(-30)}=\dfrac{1}{20}

:\implies\sf\:\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{30}

:\implies\sf\:\dfrac{1}{v}=\dfrac{3-2}{60}

:\implies\:\boxed{\bf{\gray{v=60\:cm}}}

7 0
3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
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