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3 years ago

4. How does density affect refraction of a lightwave?

Physics

1 answer:

Brums [2.3K]3 years ago

7 0

optical density would increase the refraction angle going in to the material

there may be a relation between opt density and mass density ... kg/m^3

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5. Radioactive isotopes can be used to find the age of rocks, fossils, or other artifacts. Carbon 14 has a half-life of 5,730 ye

horrorfan [7]

Answer:

Explanation:

Cuz i smart

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3 years ago

an athlete swims the distance from one end of a 50m pool to the other end in a time of 25 s what is The Athlete's average speed

vredina [299]

Answer:

2.5 N

because Average speed is equal to distance divided by time

3 0

3 years ago

10. There is less pressure when a force is present on a(n)

Nina [5.8K]

What’s the question

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3 years ago

Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time

jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

$v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)$

$\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}$

In this time, his displacement $\Delta x$ satisfies

${v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=50.4\,\mathrm m$

4 0

3 years ago

A hot air balloon is filled with 1.45 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 109 ∘ C . Wha

just olya [345]

Answer:

The volume of air after the balloon is heated = 1.95 × 10⁶ L

Explanation:

Charles Law: Charles' law states that the volume of a given mass of gas is directly proportional to the temperature in Kelvin, provided that the pressure remains constant.

It can be expressed mathematically as,

V₁/T₁ = V₂/T₂

Making V₂ The subject of the equation

V₂ = (V₁/T₁)T₂.................... Equation 1

Where V₁ = Initial Volume, T₁ = Initial Temperature, V₂ = Final Volume, T₂ = final Temperature

Given: V₁ = 1.45 × 10⁶ L, T₁ = 11 °C = (11 + 273) K = 284 K, T₂ = 109 °C = (109 + 273) = 382 K.

Substituting these values into equation 1 above,

V₂ = (1.45×10⁶)382/284

V₂ = 1.95 × 10⁶ L

Therefore the volume of air after the balloon is heated = 1.95 × 10⁶ L

6 0

3 years ago