Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.
(D)
Explanation:
The more massive an object is, the greater is the curvature that they produce on the space-time around it.
Answer:
![40h](https://tex.z-dn.net/?f=40h)
Explanation:
Use the velocity formula to solve
![v = \frac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D)
In this question, you are given velocity
, and you are given a distance,
. Time in this question is what you'll need to find.
Start by rearranging the velocity formula, to isolate for t.
![v = \frac{d}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D)
Start by multiplying both sides by t
![v(t) = \frac{d}{t}(t)\\\\vt = d](https://tex.z-dn.net/?f=v%28t%29%20%3D%20%5Cfrac%7Bd%7D%7Bt%7D%28t%29%5C%5C%5C%5Cvt%20%3D%20d)
Then divide both sides by v.
![vt\frac{1}{v} = d/v\\ \\t = \frac{d}{v}](https://tex.z-dn.net/?f=vt%5Cfrac%7B1%7D%7Bv%7D%20%3D%20d%2Fv%5C%5C%20%5C%5Ct%20%3D%20%5Cfrac%7Bd%7D%7Bv%7D)
Now that you've isolated for time, sub in your values and calculate.
![t = \frac{d}{v} = \frac{1800km}{45km/h} = 40 h](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bv%7D%20%3D%20%5Cfrac%7B1800km%7D%7B45km%2Fh%7D%20%3D%2040%20h)
Explanation:
We have,
Surface area, ![A=2\ cm^2=0.0002\ m^2](https://tex.z-dn.net/?f=A%3D2%5C%20cm%5E2%3D0.0002%5C%20m%5E2)
The current varies wrt time t as :
![q(t) = 4t^3 + 5t + 6](https://tex.z-dn.net/?f=q%28t%29%20%3D%204t%5E3%20%2B%205t%20%2B%206)
(a) At t = 2 seconds, electrical charge is given by :
![q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C](https://tex.z-dn.net/?f=q%28t%29%20%3D%204t%5E3%20%2B%205t%20%2B%206%5C%5C%5C%5Cq%282%29%20%3D%204%282%29%5E3%20%2B%205%282%29%20%2B%206%5C%5C%5C%5Cq%3D48%5C%20C)
(b) Current is given by :
![I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bdq%7D%7Bdt%7D%5C%5C%5C%5CI%3D%5Cdfrac%7Bd%284t%5E3%20%2B%205t%20%2B%206%29%7D%7Bdt%7D%5C%5C%5C%5CI%3D12t%5E2%2B5)
Instantaneous current at t = 1 s is,
![I=12(1)^2+5=17\ A](https://tex.z-dn.net/?f=I%3D12%281%29%5E2%2B5%3D17%5C%20A)
(c) Current is, ![I=12t^2+5](https://tex.z-dn.net/?f=I%3D12t%5E2%2B5)
Current density is given by electric current per unit area.
![J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2](https://tex.z-dn.net/?f=J%3D%5Cdfrac%7BI%7D%7BA%7D%5C%5C%5C%5CJ%3D%5Cdfrac%7B%2812t%5E2%2B5%29%7D%7B0.0002%7D%5C%5C%5C%5CJ%3D5000%2812t%5E2%2B5%29%5C%20A%2Fm%5E2)
Therefore, it is the required explanation.
Answer:
t = 1.62 h
Explanation:
A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.
θ_i = θ_r
If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m
tan θ₁ = x₁ / L
tan θ₁ = ![\frac{3.56}{1.87}](https://tex.z-dn.net/?f=%5Cfrac%7B3.56%7D%7B1.87%7D)
θ₁ = tan⁻¹ 1.90
θ₁ = 62.29º
for the second case x₂ = 1.46 m
tan θ₂ = x₂ / L
θ₂ = tan⁻¹ ![\frac{1.46}{1.87}](https://tex.z-dn.net/?f=%5Cfrac%7B1.46%7D%7B1.87%7D)
θ₂ = 37.98º
the difference in degree traveled is
Δθ = θ₁- θ₂
Δθ = 62.29 - 37.98
Δθ = 24.31º
as in the exercise they indicate that every 15º there is an hour
t = 24.31º (1h / 15º)
t = 1.62 h