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ser-zykov [4K]
2 years ago
8

A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t

he maximum value of acceleration? Neglect air resistance. It has the maximum value of acceleration just after it leaves the ground It has the maximum value of acceleration just before it lands on the ground Its acceleration is constant everywhere in its trajectory. It has the maximum value of acceleration halfway to the top of its trajectory. It has the maximum value of acceleration at the top of its trajectory.
Physics
1 answer:
Elodia [21]2 years ago
5 0

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

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Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?
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The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

When you look at an object located 1 light year away from you, you see it as it was 1 year ago.

If a star located 10 light years away from us suddenly brightens, or dims, or explodes, we see the event <em>10 years later.</em>

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How do you know if an experimental result is acceptable or trustworthy? What gives you confidence that your data are trustworthy
Virty [35]

Explanation:

For an experimental result to be considered acceptable, all relevant variables involved in the experiment must be taken into account, by isolating it, performing it under controlled conditions and modifying the conditions under which it takes place. This, with the objective of excluding alternative explanations in the analisis of the experimental data. Therefore, if these steps are followed appropriately, experimental data are trustworthy. The reliability of the experiment increases when it is replicated by other researchers and the same results are obtained.

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3 years ago
Read 2 more answers
A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

==> \frac{1}{2}* 550* 0.02^{2} = <em>0.11 J</em>

3 0
3 years ago
Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi
Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

I = I_0 cos^2\theta

Where,

I_0 = Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

\theta= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

I_1= \frac{I_0}{2}

In the case of the second polarizer the angle is directly 60 degrees therefore

I_2 = I_1 cos^2\theta

I_2 = (\frac{I_0}{2} ) cos^2(60)

I_2 = 0.125I_0

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

\theta_3 = 90-60 = 30

Then,

I_3 = I_2 cos^2\theta_3

I_3 = 0.125I_0 cos^2 (30)

I_3 = 0.09375I_0

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0
3 years ago
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