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3 years ago

Convert 34 kelvin to farenheit

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

7 0

Answer:

-398.47

Explanation:

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The asteroid ceres lies at an average distance of 414 million kilometers from the sun. The period of revolution of ceres around

7nadin3 [17]

Answer:

T2 = 1680,4 days

Explanation:

Kepplers law:

$\frac{T^{2} }{a^{3} } = constant$

For Earth:

T1 = 365 days ; a1 = 149 597 870 700 m

For Ceres:

T2 = ? days ; a1 = $414*10^{9} m$

Then:

$\frac{T1^{2} }{a1^{3} } = \frac{T2^{2} }{a2^{3} } ----> T2 = T1*\sqrt{\frac{a2^{3}}{a1^{3}}}$

Replacing values:

T2 = 1680,4 days

6 0

3 years ago

Read 2 more answers

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Rom4ik [11]

Answer: $Q=0.5612 m^3/s$

Explanation:

Given

diameter of pipe( $d_1$ )=6 cm

diameter of pipe( $d_2$ )=4 cm

$P_1=32 kPa$

$P_2=24 kPa$

$A_1=\frac{\pi }{4}6^2=9\pi cm^2$

$A_2=\frac{\pi }{4}4^2=4\pi cm^2$

$v_1=\frac{Q}{A_1}$

Applying bernoulli's equation

$\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2$

$\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2$

since $z_1=z_2$

$\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}$

$Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}$

$Q^2=3149.3722\times 10^{-4}$

$Q=\sqrt{3149.3722\times 10^{-4}}$

$Q=0.5612 m^3/s$

4 0

3 years ago

A light-year measures the _______ that light travels in 1 year.

o-na [289]

<span>A light-year measures the distance that light travels in 1 year.

Answer : B ) Distance

-Hope this helps.</span>

4 0

4 years ago

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Suppose that a block is pulled 16 meters across a floor. What amount of work is done if the force used to drag the block is 22 N

klemol [59]

We have: A= F.s
So, A = 22 x 16
A= 352J
Só A is correct answer

3 0

3 years ago

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Which option below correctly compares the average annual dose of background radiation to the dose liked to an increased cancer r

julia-pushkina [17]

The answer for this question would be choice "<span>B. The average annual dose of background radiation is 250 times smaller than the dose linked to increased cancer risk."

You only have to compare 4.0 x 10^-4 and 1.0 x 10^-1. And if you can observe carefully, when you try to multiply the average annual dose of background radiation by 250, you would get 0.1 which is equivalent to the amount of annual dose linked to increased cancer risk. Therefore, the answer is B.</span>

6 0

3 years ago