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Dmitry [639]
4 years ago
8

An engine manufacturer makes the claim that the engine they have developed will, on each cycle, take 100J of heat out of boiling

water at 100∘C, do mechanical work of 80J, and exhaust 20J of heat at 10∘C. What, if anything, is wrong with this claim?
Physics
1 answer:
Alborosie4 years ago
4 0

Answer:

Explanation:

Intake heat, QH = 100 J

output heat, Qc = 20 J

Work, W = 80 J

TH = 100°C = 373 K

Tc = 10°C = 283 K

TH/ Tc =  373 / 283 = 1.318

QH/Qc = 100 / 20 = 5

for a heat engine, those ratios should be same. so temperature is not correct.

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<span>The factors that determine how fast weathering occurs are the type of rock, type of soil, time it takes, and the climate.</span>
5 0
4 years ago
Which object has the greatest amount of kinetic energy?
EleoNora [17]
I think that the answer is A
4 0
3 years ago
The practical limit to an electric field in air is about 3.00 × 10^6 N/C . Above this strength, sparking takes place because air
AleksAgata [21]

Answer:

(a) x=157 m

(b) No

Explanation:

Given Data

Mass of proton m=1.67×10⁻²⁷kg

Charge of proton e=1.6×10⁻¹⁹C

Electric field E=3.00×10⁶ N/C

Speed of light c=3×10⁸ m/s

For part (a) distance would proton travel

Apply the third equation of motion

(v_{f})^{2} =(v_{i})^{2}+2ax

In this case vi=0 m/s and vf=c

so

c^{2}=(0)^{2}+2ax\\  c^{2}=2ax\\x=\frac{c^{2} }{2a}

x=\frac{c^{2}}{2a}--------Equation (i)

From the electric force on proton

F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\

put this a(acceleration) in Equation (i)

So

x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2}  }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m

For part (b)

No the proton would collide with air molecule

7 0
3 years ago
Find the magnitude of the sum of these two vectors:​
sukhopar [10]

Answer:

Explanation:

You can decompose those vectors into their components in x and y direction. For the first vector:

\vec{r}_{1}=r_{1}\cos 30\hat{i}+r_{1}\sin 30\hat{j}=3.14(\frac{1}{2}\sqrt{3}\hat{i}+\frac{1}{2}\hat{j})=1.57\sqrt{3}\hat{i}+1.57\hat{j}

For the second vector:

\vec{r}_{2}=r_{2}\cos 60\hat{i}-r_{2}\sin60 \hat{j}=1.355\hat{i}-1.355\sqrt{3}\hat{j}

The sum of two vectors will be:

\vec{r}=\vec{r}_{1}+\vec{r}_{2}=(1.57\sqrt{3}+1.355)\hat{i}+(1.57-1.355\sqrt{3})\hat{j}\approx 4.0711\hat{i}-0.77415\hat{j}

The magnitude of the sum of two vectors is:

r=\sqrt{(4.0711)^{2}+(-0.77415)^{2}}\approx 4.14 meter

7 0
2 years ago
Using a refracting telescope, you observe the planet Mars when it is 1.95 × 10 11 m from Earth. The diameter of the telescope's
liubo4ka [24]

Answer:

The minimum feature size is 136.6 km.

Explanation:

It is given that,

Distance from planet Mars and Earth is, D=1.95\times 10^{11}\ m

Diameter of the objective lens, d = 0.977 m

We need to find the minimum feature size, on the surface of Mars that your telescope can resolve for you. The expression for resolving distance is given by :

y=\dfrac{1.22D\lambda}{d}\\\\y=\dfrac{1.22\times 1.95\times 10^{11}\times 561\times 10^{-9}}{0.977 }\\\\y=136603.78\ m\\\\y=136.6\ km

So, the minimum feature size is 136.6 km.

8 0
3 years ago
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