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2 years ago

The critical angle for water is 49°. If a ray of light

Physics

1 answer:

Sonja [21]2 years ago

3 0

Answer:

Snell's Law states

Ni sin i = Nr sin r

Judging from the question the source of the ray is in the water (directed up)

or NI = 1 / sin 49 Ni = 1.325 deg the critical angle

From inside the pond:

Nr = 1.325 * sin 45 / 1 = 94 deg

So refraction can occur outside the pond and you do not have total internal refection.

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A sample of gas in a syringe releases heat to its surroundings while the

blsea [12.9K]

Answer: b

Explanation:

When heat is released by the system i.e. system loses heat. So, we take it as negative -Q

When the work is done on the system then it is considered as negative work on the system i.e. -W

In this case, the plunger is pulled out, and work is done on the system. So, we take work as negative work -W

Correct option is b

3 0

2 years ago

Read 2 more answers

An unknown material has a mass

atroni [7]

Answer: 1896.55J/kg°C

Explanation:

The quantity of Heat Energy (Q) required to heat a material depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = 1320 joules

Mass of material = 5.61kg

C = ? (let unknown value be Z)

Φ = 0.124°C

Then, Q = MCΦ

1320J = 5.61kg x Z x 0.124°C

1320J = 0.696kg°C x Z

Z = (1320J / 0.696kg°C)

Z = 1896.55 J/kg°C

Thus, the specific heat of the material is 1896.55J/kg°C

4 0

3 years ago

Read 2 more answers

A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold

salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o

devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒ $=\frac{280\times (3)^2}{4 \pi^2}$

⇒ $=\frac{280\times 9}{4\times (3.14)^2}$

⇒ $=68.83 \ kg$

(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

⇒ $=0.725 \ m/s$

4 0

2 years ago

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

kirill115 [55]

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

8 0

2 years ago