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3 years ago

Will the sun ever stop shining? If so, how and when would we know((there are 3 questions

Physics

1 answer:

Alinara [238K]3 years ago

6 0

Answer:

Explanation:stop it cause the sun don't shine that long unless if it's summer cause the summer time is very hot it may be 100 degrees or lower

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A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met

zlopas [31]

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

-kx - bv = ma

-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

$D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}$

For an overdamped system,

$(b/m)^{2} - 4k/m} > 0$

$(b/m)^{2} > 4k/m}\\(b/m) > \sqrt{4k/m}} \\(b/m) > 2\sqrt{k/m}} \\b > 2\sqrt{km}}$

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

6 0

2 years ago

A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the

erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, $T_{b} = 75^{\circ}C$

Air temperature, $T_{infty} = 20^{\circ}$

k = 388 W/mK

h = $20\ W/m^{2}K$

Now,

Perimeter of the fin, p = $\pi D = 0.005\pi \ m$

Cross-sectional area of the fin, A = $\frac{\pi}{4}D^{2}$

A = $\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}$

To calculate the heat transfer rate:

$Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})$

where

$m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237$

Now,

$Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W$

5 0

3 years ago

Which of The following is the best example of water changing from a liquid to gas

expeople1 [14]

Answer:

There are no examples but this should be evaporation

Explanation:

4 0

3 years ago

How do you calculate the volume and density of something that weighs 20 grams?

Nostrana [21]

Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
i believe that's the answer.
oh gosh i didn't realize my middle school education involved high school stuff.

6 0

3 years ago

I don’t get this question and i need help!!!

galina1969 [7]

i would say number 2

3 0

3 years ago