The given data is incomplete. The complete question is as follows.
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)
Explanation:
Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and 
be the friction coefficient and m be the mass of car.
Hence, the given data is as follows.
v = 0, s = 84 m, = 0.36
According to Newton's law of second motion the expression for acceleration is as follows.
F = ma
= ma
= ma
a = 
Also,
= 
= 24.36 m/s
Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.
A) be too hot to support life
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
Symbols such as cross motivates me to go to church.
Answer:
1.5 x 10⁵ W
Explanation:
A = Area of the fresh lava = 1.02 m²
T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K
T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K
ε = emissivity of the lava = 0.97
σ = stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴
Rate of transfer is given as
E = σ ε A (T⁴ - T₀⁴)
E = (5.67 x 10⁻⁸) (0.97) (1.02) ((1273)⁴ - (298.3)⁴)
E = 1.5 x 10⁵ W
