The sloping surface of the inclined plane<span> supports part of the weight of the object as it moves up the slope. As a result, it takes less </span>force<span> to move the object uphill. The </span>trade<span>-off </span>is<span> that the object must be moved over a greater </span>distance<span> than if it were moved straight up to the higher elevation.</span>
Answer:
See the explanation below.
Explanation:
The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.
s = displacement [m]
v and u = velocity [m/s]
t = time [s]
Now using these units in the given equation.
![s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]](https://tex.z-dn.net/?f=s%20%3D%200.5%2A%28%5Bm%2Fs%5D%2B%5Bm%2Fs%5D%29%2A%5Bs%5D%5C%5Cs%3D0.5%2A%5Bm%2Fs%5D%2A%5Bs%5D%5C%5Cs%20%3D%200.5%2A%5Bm%5D)
So the expression is good, and dimensional has consistency.
Answer:
That would be a perpetual motion machine
Explanation:
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
The correct answer for this question is this one: "measuring the temperature increase of water from doing work stirring it." This experiment is generally regarded as being first carried out by James Joule is this one, <span>measuring the temperature increase of water from doing work stirring it.</span>
