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Nikitich [7]
3 years ago
15

Answer with explanation!!!

Physics
1 answer:
adell [148]3 years ago
8 0

Answer:

The light enters the box along the normal to the side of the box or perpendicularly to the box's surface

Therefore, the light is expected to travel straight through the box

However, when a rectangular glass prism is placed inside the box, the light can then be refracted to pass through the box in the given path as shown

Light bends towards the normal when passing from a less dense medium to a denser medium with larger refractive index and vice versa

Therefore, when the glass prism with a larger refractive index than air is inclined with the top part further away from the incident beam and the bottom part closer to the incident beam as shown, the refracted ray through the box will be shifted downwards as shown in the drawing created with Microsoft Visio

Explanation:

Refractive \ index, n = \dfrac{sin(i)}{sin(r)}

Where;

i = The angle between the incidence light and the normal line

r = The angle between the refracted light and the normal line

n = The refractive index

When n > 1 is large, we have, ∠i > ∠r and the light is bent towards the normal when moving from a less dense medium to a denser medium and vice versa.

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Mv + mv = 2mv providing each momentum is in the same direction.
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4 0
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A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a
butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance C=260\ pF

Charge q=0.155\ \mu\ C

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

V= \dfrac{Q}{C}

Where, Q = charge

C = capacitance

Put the value into the formula

V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}

V=596.2\ volts

Hence,The potential difference between the plates is 596.2 volts.

7 0
3 years ago
9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en
masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

P_1 = Presión inicial = 0.96 atm

P_2 = Presión final

V_1 = Volumen inicial = 95 mL

V_2 = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0
2 years ago
A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
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Answer:

182.28 W

Explanation:

Here ,

m = 7.30 Kg

distance , d= 28.0 m

time , t = 11.0 s

average power supplied = change in potential energy/time

average power supplied = m×g×d/time

average power supplied = 7.30×9.81×28/11

average power supplied = 182.28 W

the average power supplied is  182.28 W

6 0
2 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

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\lambda=4.8 m

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(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

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A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

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