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4 years ago

Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?

2 answers:

6 0

Explain why it is not advisable to use small values of incident ray in performing experiment on therefraction through a glass prism.

Colt1911 [192]4 years ago

3 0

Answer:

Explanation:

Prism is a glass object that is used to study the pattern of refraction.

During this experiment it is not advised to used small values of incident angle , as when a small amount of incident ray is passed the refracted pattern observed at the end will not provide satisfactory results.

The amount of light that comes out of prism which is the refracted light will also be small hence study of complete refracted pattern will not be possible.

The small incident angle will give rise to more errors in measurement.

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

3 years ago

A block of mass 20 kg sits on a ramp with an angle of 31 degrees above the horizontal. Assuming no friction, how fast will the b

Artyom0805 [142]

Answer:

2.98 m/s^2

Explanation:

I have done this before and it was a question on my physics test

4 0

4 years ago

Read 2 more answers

Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when s

lubasha [3.4K]

Answer:

Explanation:

Let the internal resistance be r .

Since in open circuit the volt is 1.55 V , this will be the source voltage .

Source voltage = 1.55

If external resistance be R .

1.55 / (R + r ) = .500

R + r = 3.1 ohm

So sum of internal resistance and external resistance will be 3.1 ohm.

7 0

3 years ago

Where did the missouri compromise imaginary line run

lara31 [8.8K]

In the Missouri Compromise, the slavery line for future US states ran along the southern border of Missouri at 36 degrees north 30 minutes

7 0

3 years ago

Longitudinal sound waves cannot propagate through

strojnjashka [21]

Answer:

A vacuum

Explanation:

Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.

For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.

Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:

a vacuum

represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.

3 0

3 years ago