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4 years ago

When leaving a fish tank, a ray moving at 31.0 deg in glass (n = 1.50) exits into the air. At what angle does it enter the air?

Physics

1 answer:

stiv31 [10]4 years ago

8 0

Answer: 20 degrees

Explanation:

Given that the

Refractive index n = 1.5

Incident angle = 31 degree

The refractive index formula is:

n = sin I/sin R

1.5 = sin 31/ sin R

Make sin R the subject of formula

Sin R = sin 31 / 1.5

Sin R = 0.515 / 1.5

Sin R = 0.343

R = sin^-1(0.34335)

R = 20 degrees

Therefore, it enters the air at angle 20 degrees

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Nonamiya [84]

Answer:

Action - Pulling up the train.

Reaction - Friction on the locomotive

Explanation:

Locomotive is pulling the train upwards ,

Which is the action force applied by the locomotive,

As a reaction locomotive will be pulled by the train which is the reaction of pulling

Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.

For train action is pulling up by locomotive and reaction will be friction acting on it downwards.

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3 years ago

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3 years ago

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Answer:

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3 years ago

A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the

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Answer:

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3 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

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3 years ago