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baherus [9]
3 years ago
6

HELP ME the bending of light when it changes media is called _________ u can get brainliest

Physics
2 answers:
olga55 [171]3 years ago
6 0

Answer:

Refraction

Explanation:

Arte-miy333 [17]3 years ago
5 0
(In my opinion) The bending of light when it changes media is called REFRACTION. good luck!
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Verdich [7]

when you carry the 2 and add the 3 its Z

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3 years ago
A complete digestive system is found in bears.
GrogVix [38]
Hi! I believe the correct answer is A. Do you have any other questions I can help you with? I would be glad too.

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4 0
3 years ago
Quickly spinning the handle of a hand generator, Kristina is able to light three bulbs in a circuit. When she spins the generato
erastova [34]

Answer:

  • <u><em>Voltage is too low</em></u>

Explanation:

<em>Generators</em> produce a potential difference (voltage) by <em>spinning </em>a coil of wire in a magnetic field by induction.

The  produced voltage is proportional to the relative speed of the coil wire with respect to the magnetic field.

Hence, for a given generator, the higher the speed the higher the potential difference.

Therefore, it is concluded immediately that when Kristina spins the generator slowly, less potential difference (voltage) is generated.

Also, you must know that the brightness of the bulbs varies with the current: the higher the current, the brigther the bulbs.

On the other hand, the basic realtion between voltage (V), resistance (R), and current I) is given by Ohm's law: V = R × I. This is, voltage and current are proportional.

Since the generator is producting less voltage, for the same bulbs (which are resistors) , the current will be lower too. And, as stated, lower current means lower brightness, which explains why when Kristina spins the generator slowly, the bulbs are very dim.

5 0
3 years ago
g If 6.35 moles of a monatomic ideal gas at a temperature of 320 K are expanded isothermally from a volume of 1.45 L to a volume
vitfil [10]

Answer:

(a)  change in the internal energy of the gas is zero

(b) the work done by the gas is 16.93 kJ

(c) the heat flow is 16.93 kJ, which is into the gas

Explanation:

Given;

number of moles of gas, n = 6.35 moles

temperature of the gas, T = 320 K

initial volume of the gas, V₁ = 1.45 L

final volume of the gas, V₂ = 3.95 L

Part (a)

For isothermal expansion, temperature is constant and internal energy will also be constant.

Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).

ΔU = Q - W

where;

ΔU is change in internal energy

Q is heat transferred to the system

W is the work done by the system

Thus, Q = W

ΔU = 0

Part (b)

the work done by the gas

W = nRTln{[\frac{V_2}{V_1}]

where;

R is gas constant = 8.314 J/mol.K

W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ

Part (c)

the heat flow into or out of the gas

Q = ΔU + W

Q = 0 + 16.93 kJ

Q = 16.93 kJ

Since the heat flow is positive, then it is heat flow into the gas.

3 0
3 years ago
n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we
USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

        p₀ = m v₁ + M 0

final instant. Right after the crash

        p_f = (m + M) v

        p₀ = p_f

        mv₁ = (m + M) v

        v = \frac{m}{m+M} \  v_1

     

we substitute

        v = \frac{20}{20+40}   3

        v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

        K₀ = ½ m v₁² + 0

        K₀ = ½ 20 10³ 3²

        K₀ = 9 10⁴ J

        K_f = ½ (m + M) v²

        K_f = ½ (20 +40) 10³  1²

        K_f = 3 10⁴ J

the change in energy is

       ΔK = K_f - K₀

       ΔK = (3 - 9) 10⁴

       ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0
3 years ago
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