Answer:
Thermal energy flows from a warmer object to a cooler object.
Explanation:
The second law of thermodynamics speaks about the direction of processes. Clarifying that these can only go in one direction as time progresses in one direction. To better understand this concept, several examples will be explained.
The first example is when we have a wood that we will use to make a campfire, as we require thermal energy to heat a space. We see that as the fire consumes the wood heat is generated inside the space. This will happen until the wood is consumed. Now if we want to perform the reverse process, that is to cool the room and use the heat content in this to generate wood, we see that it is an impossible process to perform, that is to say, it is true that this thermal process has only one direction.
Now suppose we have a cup of coffee at a temperature of 95 (°C), with the passage of time the coffee cools this is because it generates a heat transfer from the coffee to the surroundings. Similarly if we want to collect the heat from the surroundings and pass them to the coffee so that it reaches a temperature close to 95 (°C), naturally it can not be done, the heat transfer always exists in one direction this is from objects at high temperature to objects of lower temperature.
To violate the second law of thermodynamics, and to allow processes to go in the opposite direction, it is only possible through mechanical work that adheres to the process. For example if we have a canned drink at 25 (°C) and want to reduce the temperature of this drink, then we must put it in a refrigerator, so that it can reduce its temperature to 5 (°C). The refrigerator is electromechanical equipment that can alter heat transfer directions.
Tapping the brakes is almost definitely going to be better than not tapping the brakes.
Using the brakes causes the static friction between your tires and the road to increase; which slows down the car more.
The issue is that when the grip between the brakes and the tire is too strong, the coefficient of friction between the tires isn't high enough to maintain that static contact. This is when the wheels "lock up" and begin to skid. As you mention, kinetic friction is less than static friction, so wheel lock and skidding should be avoided.
This means that tapping the brakes is generally the best braking method in these conditions, as compared to not braking at all.
When considering modern vehicles, your friend missed the mark even more. Most vehicles sold now (as far as I know) have an ABS, an Anti-lock Braking System which actually disengages the brakes when they reach the threshold when you start to lose traction. Basically, you can push down on the brake pedal as hard as you want, and you can feel the brakes start "tapping" on their own, to prevent skidding, increase traction, and allow you to brake in the lowest space possible. Tapping is the method most cars automatically employ on icy surfaces, and if you try it for yourself (in a controlled environment), you will find that tapping the brakes should slow you down a lot faster than not pressing it at all; especially if you avoid skidding.
Hence,
Tapping the brakes is almost definitely going to be better than not tapping the brakes.
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To determine the amplitude of a wave, you would need to know the : B. Maximum displacement of the wave from the equilibrium position
You can find out the displacement by counting the waves that is
written on the Graph
Hope this helps
Answer:
y = -0.42 m
1.16*10^-4 T(-k)
-1.73*10^4 N/C (j)
Explanation:
(a) Above the pair of wires, The field out of the page of the 50 A current will be stronger than the (—k) field of the 30 A current (k).
Between the wires, both produce fields into the page.
below the wires, y = - | y |
B = u_o*I/2πr (-k) + u_o*I/2πr (k)
0 = u_o/2πr[50/ | y |+0.28 (-k) + 30/| y | (k) ]
50 | y | = 30(| y | + 0.28)
| y | = -y
-50 y = 30*(0.28 - y)
y = -0.42 m
b) B = u_o*I/2πr (-k) + u_o*I/2πr (k)
B = 4π*10^-7/2π[ 50/0.28 -1 (-k) +30/1(-k) ]
= 1.16*10^-4 T(-k)
F = qv*B
F = (-2*10^-6)*(150*10^6(i) )(1.16*10^-4(-k))
F = 3.47*10^-2 N(-j)
c) F_e = qE
E = F_e/q
E = 3.47*10^-2/-2*10-6
= -1.73*10^4 N/C (j)
