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Georgia [21]
3 years ago
14

Care este pozitia unui mobil cand t(s)=14 si x(m)=12?​

Physics
1 answer:
mario62 [17]3 years ago
3 0

name shrawan sha

DtD_LpOZ3Xm_0ff

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What happens to the speed of the particles if the temperature is increased
dem82 [27]
In a simplistic model, the average speed of the particles increases when temperature is increased.
8 0
4 years ago
What formula could be used to find distance if you know the speed an the time
gogolik [260]

Answer: d = st

Explanation:

We know that the distance is equal to the rate (speed) times the time

d = st

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3 years ago
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Answer:

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4 0
3 years ago
If the plant produces electric energy at the rate of 1. 5 gw , how much exhaust heat is discharged per hour?
expeople1 [14]

The amount of exhaust heat discharged per hour from the plant which produces electric energy at the rate of 1. 5 gw, is 9.36×10¹⁹ J/h.

<h3>How to calculate the heat discharged per hour?</h3>

The heat is discharged per hour is equal to the different of total power and actual power.

The plant produces electric energy at the rate of 1. 5 gw. Let assume the efficiency of this power plant is 64% between the temperature of 660 degree C to 330 degree Celsius.

Thus, the maximum efficiency is,

\eta_{max}=1-\dfrac{660+273}{330+273}\\\eta_{max}=0.55

The total power is,

P_t=\dfrac{P_A}{\eta_{max}\times0.64}\\P_t=\dfrac{1.5}{0.55\times0.64}\\P_t=4.26\rm\; GW

Thus, the heat discharged per hour is,

Q=4.26-1.5\\Q=2.76\text{GW}\\

Multiply the value with 3600 to convert it in s/h,

Q=2.76\times10^9\times3600\text{ J/h}\\Q=9.36\times10^{19}\text{ J/h}\\

Thus, the amount of exhaust heat discharged per hour from the plant which produces electric energy at the rate of 1. 5 gw, is 9.36×10¹⁹ J/h.

Learn more about the heat discharged per hour here;

brainly.com/question/19666326

#SPJ4

6 0
2 years ago
What is the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 k
sesenic [268]

Answer:

the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

Explanation:

Using Rayleigh criterion for the limiting angle of resolution of an eye

\theta = \frac{1.22\lambda }{D } \\ \\  \theta = \frac{1.22*506 *10^{-9} }{4.90*10^{-3}m}

\theta = 1.2598*10^{-4} rad

\theta = 125.98*10^{-6} \ rad

Thus; the separation  between the two sources is expressed as:

\theta = \frac{y}{L} \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^{-6} \ rad) \\ \\ y = 31.495 \ m

Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

8 0
3 years ago
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