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3241004551 [841]
3 years ago
13

When a falling object reaches terminal velocity, the net force acting on it is

Physics
1 answer:
dexar [7]3 years ago
3 0
Pretty sure that it is 0.
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The rest deltoid row is a back exercise true or false
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False because your deltoids are in your shoulders not your back
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3 years ago
The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p
Blizzard [7]

Answer:

True

Explanation:

In this particular case, the area of the graph represents the impulse.

In fact, impulse is defined as the change in momentum of an object:

I=\Delta p

Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:

I=F\Delta t

If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product

F\cdot \Delta t

which corresponds to the definition of impulse.

8 0
3 years ago
An object weighs 42.53 newtons. What is its mass if a gravitometer indicates that g = 9.83 m/s2? Round to the nearest tenth.
enot [183]
42.53/9.83 = 4.3265.... = 4.3
3 0
3 years ago
Read 2 more answers
How is a reference point defined?
nevsk [136]
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6 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
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