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Svetradugi [14.3K]
3 years ago
12

A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 N. 825 N. frictional force a

cting on the box as it moves to the right what is the net force in the Y direction
Physics
1 answer:
lidiya [134]3 years ago
8 0

Answer:A

Explanation:

Explanation:

Given that,

Gravitational force = 600 N

Frictional force = 25 N

Pulled by the Force = 250 N

We know that,

The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.

The balance equation is along y-axis

The box will not move in y-axis therefore, the net force in the y-axis will be zero.

Hence, The net force in the y-direction will be zero.

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What would the color of this star cluster appear to be if you observed it from a distance so great that you could not make out t
makkiz [27]

Answer:

D. blue

Explanation:

4 0
3 years ago
An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to
Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

\theta_1 is the angle of incidence (in the first medium)

\theta_2 is the angle of refraction (in the second medium)

Let's now consider a situation in which

n_1 > n_2

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

Where \frac{n_1}{n_2} is a number greater than 1. This means that above a certain value of the angle of incidence \theta_1, the term on the right can become greater than 1. So this would mean

sin \theta_2 > 1

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

\theta_c =sin^{-1}(\frac{n_2}{n_1})

8 0
3 years ago
While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an ident
Feliz [49]

Answer:

-6.49 m/s

Explanation:

This is doppler effect.

The equation is;

F_l = [(v + v_l)/(v + v_s)]F_s

Where;

F_l is frequency observed by the listener

v is speed of sound

v_l is speed of listener

v_s is speed of source of the sound

F_s is frequency of the source of the sound

In this question, the source of the sound is the moving vehicle.

Thus;

F_l = F_beat + F_s

We are given beat frequency (f_beat) as 5 Hz while source frequency (F_s) as 260 Hz.

So,

F_l = 5 + 260

F_l = 265 Hz

Since listener is sitting by car, thus; v_l = 0 m/s

Thus,from our doppler effect equation, let's make v_s the subject;

v_s = F_s[(v + v_l)/F_l] - v

Speed of sound has a value of v = 344 m/s

Thus;

v_s = 260[(344 + 0)/265] - 344

v_s = -6.49 m/s

This value is negative because the source is moving towards the listener

3 0
3 years ago
If a blue whale weighs 200 tons what would it be on the moon?
sasho [114]

Answer:

if no mistaken then it is 300kN

Explanation:

Using SI units (the default if a system is not specified) 200T = 200000kg as the mass of the whale, it would weigh 1960kN in 1g Earth gravity.

On the Moon, the acceleration is 0.166g, to it would weigh about 330kN.

7 0
2 years ago
Read 2 more answers
11. A student lifts a 25 kg mass a vertical distance of 1.6 m in a time of 2.0 seconds.
goldenfox [79]

Answer:

a) 245 N

b) 392 J

c) 196 W

Explanation:

a) Force needed to lift the 25 kg object is the same as the weight of the object:

Force= m * g = 25kg * 9.8 \frac{m}{s^2} = 245 N

b) Work is the force exerted times the distance moved:

Work = F * d = 245N * 1.6 m = 392 Joules = 392 J

c) The power exerted is the quotient of the work done over the amount of time it took to move the object:

4 0
3 years ago
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