What should you do if you get caught in quicksand
1 answer:
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Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
![T = \sqrt[4]{202.857 * 10 ^{8} }](https://tex.z-dn.net/?f=T%20%3D%20%20%5Csqrt%5B4%5D%7B202.857%20%2A%2010%20%5E%7B8%7D%20%7D%20)
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K.
Answer:
The radius is
Explanation:
From the question we are told that
The distance beneath the liquid is
The refractive index of the liquid is
Now the critical value is mathematically represented as
substituting values
Using SOHCAHTOA rule we have that
=>
substituting values
Answer:
t₂ = 3.89 s
Explanation:
given,
speed of car = 23 m/s
speed of motorcycle = 23 m/s
after time of 4 s distance between them is equal to = 53 m
motorcycle accelerates at = 7 m/s
time taken to catch up with car = ?
let t₂ be the time in which motorcycle catches car.
distance traveled by car in t₂ s
d = 23 t₂ + 53
distance traveled by motorcycle
using equation of motion
now, equating both the distances
t₂ = 3.89 s
time taken by the motorcycle to catch the car is equal to 3.89 s