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3 years ago

Astronomers think most galaxy interactions took place at redshifts of greater than 1 because

Physics

1 answer:

Evgesh-ka [11]3 years ago

4 0

Because they are moving away from us.

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A car travels up a hill at a constant speed of 22 km/h and returns down the hill at a constant speed of 55 km/h. calculate the a

Ket [755]

The answer choice is 2.5

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3 years ago

PLEASE ANSWER USING ALL MY POINTS FOR THIS NEED HELP (please answer all of them)

kipiarov [429]

4-C. ... 5-D. ... 6-A. ... 7-D. ... 8-C. ... 9-B. ... 10-D. It's really a joy and a delight that you've learned so much by posting all of these questions.

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4 years ago

How much heat is absorbed from a 56.00 g sample of Mercury when its tempreature change is 289K?

AveGali [126]

Heat required to raise the temperature of mercury is given as

$Q = ms\Delta T$

here given that

m = 56 g

s = specific heat capacity of mercury = 0.140 J/g C

$\Deta T = 289 k$

now here we have

$Q = 56 * 0.140 * 289$

$Q = 2265.76 J$

so it required 2265.76 J of heat

8 0

3 years ago

A barge ﬂoating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H

Karolina [17]

Answer:

a) $\Delta m = 330000\,kg$ , b) $h = 1.268\,m$

Explanation:

a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:

$\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V$

$\Delta m = \rho_{w}\cdot \Delta V$

$\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)$

$\Delta m = 330000\,kg$

b) The submersion height is found by using the equation derived previously:

$\Delta m = \rho_{w}\cdot \Delta V$

$450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)$

The final submersion height is:

$h = 1.268\,m$

4 0

3 years ago

Read 2 more answers

A 200 kg roller coaster is located at the top of a hill where the height is 30 m. What is the velocity when it reches the bottom

maw [93]

Assuming it starts at rest, the roller coaster only has potential energy at the top of the hill, which is

$E_P=(200\,\mathrm{kg})(30\,\mathrm m)g$

When it reaches the bottom, its potential energy will have converted to kinetic energy,

$E_K=\dfrac12(200\,\mathrm{kg})v^2$

where $v$ is its velocity at that point. By the law of conservation of energy, assuming no loss of energy due to other sources (e.g. sound, heat), we have

$E_P=E_K\iff(6000\,\mathrm{kg}\cdot\mathrm m)g=(100\,\mathrm kg})v^2$

$\implies v=\sqrt{60g}\,\dfrac{\rm m}{\rm s}\approx24.2\,\dfrac{\rm m}{\rm s}$

8 0

4 years ago

Read 2 more answers