When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).
The value of this critical angle can be derived by Snell's law, and it is equal to
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In our problem, n1=1.47 and n2=1.33, so the critical angle is
By definition of average acceleration,
<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²
Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:
∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N
If you just want the magnitude, drop the negative sign.
You must run at least 53.3333333 meters a minute.
Answer:
Decreases to half.
Explanation:
From the question given above, the following data were obtained:
Initial mass (m₁) = m
Initial force (F₁) = F
Initial acceleration (a₁) =?
Final mass (m₂) = ½m
Final force (F₂) = ¼F
Final acceleration (a₂) =?
Next, we shall determine a₁. This can be obtained as follow:
F₁ = m₁a₁
F = ma₁
Divide both side by m
a₁ = F / m
Next, we shall determine a₂.
F₂ = m₂a₂
¼F = ½ma₂
2F = 4ma₂
Divide both side by 4m
a₂ = 2F / 4m
a₂ = F / 2m
Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:
a₁ = F / m
a₂ = F / 2m
a₂ : a₁ = a₂ / a₁
a₂ / a₁ = F/2m ÷ F/m
a₂ / a₁ = F/2m × m/F
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
From the illustrations made above, the acceleration of the car will decrease to half the original acceleration
