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4 years ago

Convert the temperature 288 K to degrees Celsius

2 answers:

6 0

Answer: 14.85°C

Justification:

1) To convert Kelvin temperature tu Celsius temperature you subtract 273.15

2) This is the formula.

°C = K - 273.15

3) The conversion of 288 K to degrees Celsius is:

°C = 288 - 273.15 = 14.85 K

Kelvin temperature is called absolute temperature. The lower bound of the absolute temperature is 0 K.<span>, that means absolute zero and it is physically impossible to reach.

At 0 K or absolute temperature the atoms would stop vibrating meining that the entropy of the substance reaches it's minimum value, i.e. 0

</span>Zero K is equivalent to - 273.15°C

Bumek [7]4 years ago

6 0

i dont know how to convert it but you can use goolge to transfer it

288 kelvin = 14.85 celsius

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.89 cm2 is held fixed at the end points of the rod. What is

Vedmedyk [2.9K]

Answer:

The size of the force developing inside the steel rod is 32039.28 N

Explanation:

Given;

length of the steel rod, L = 1.55 m

cross sectional area of the steel, A = 4.89 cm²

temperature change, ∆T = 28.0 K

coefficient of linear expansion for steel, α = 1.17 × 10⁻⁵ 1/K

Young modulus of steel, E = 200.0 GPa.

Extension of the steel is given as;

α ∆T L = FL / AE

α ∆T = F/AE

F = AEα ∆T

F = ( 4.89 x 10⁻⁴)(200 x 10⁹)(1.17 × 10⁻⁵)(28.0 K)

F = 32039.28 N

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3 years ago

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Troyanec [42]

Answer:

(a) Radius of orbit will be $=7.32\times10^6m$

(b) Earth gravitational force will be $=4.18\times 10^4N$

Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

Time period T = 6200 sec

We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

(a) We know that $\omega ^2=\frac{GM}{R^3}$

$R^3=\frac{GM}{\omega ^2}$

$R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}$

$R^3=3.92\times 10^{20}$

Radius of the orbit $R=7.32\times 10^6m$

(b)

Force $F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N$

(c)

Altitude $h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m$

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3 years ago

a crowbar having the lenght of 1.75 is used to balance a load of 500N if the distance between the fulcrum and the load is 0.5m f

goblinko [34]

The effort applied : 200 N

<h3>Further explanation</h3>

Equilibrium rotation

$\large {\boxed {\bold {\sum \tau = 0}}$

F₁.d₁-F₂.d₂=0

F₁.d₁=F₂.d₂

F₁=500 N

d₁=0.5 m

d₂=1.75-0.5=1.25 m

$\tt 500\times 0.5=F_2\times 1.25\\\\F_2=\dfrac{500\times 0.5}{1.25}=200~N$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Maksim231197 [3]

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