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3 years ago

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Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

he more distant a galaxy is, the younger it appears. C. The older the galaxy appears to us, the more luminous it is. D. The faster the galaxy spins, the more massive and luminous it is. E. The greater the distance to a galaxy, the fainter it is.

1 answer:

olya-2409 [2.1K]3 years ago

4 0

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

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A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object

Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg

Explanation:

We have given mass of the first object $m_1=3.5kg$ and its velocity $v_1=8.1m/sec$

Mass of the second object $m_2=M$ it is at rest so its velocity $v_2=0$

From conservation of momentum we know that

Initial momentum = final momentum

So $m_1v_1+m_2v_2=(m_1+m_2)v$

$3.5\times 8.1+M\times 0=(3.5+M)5.2$

$28.35=18.2+5.2M$

M = 1.951 kg

8 0

3 years ago

A man pulls a 10 kg box at constant speed across the floor with a rope. The tension in the rope is 120 N at an angle of 30°.

Murrr4er [49]

Answer:

98n

Explanation:

5 0

2 years ago

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

stepan [7]

The object rises to a height of 20.4 m. So option C is correct.

Explanation:

initial velocity= Vi=20 m/s

final velocity at the top= Vf=0

acceleration= g=-9.8 m/s²

Vf²=Vi²+2gh

0= (20)²+2 (-9.8)h

-200=-9.8h

h=200/9.8

h=20.4 m

Thus the object rises to a height of 20.4 m

7 0

3 years ago

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba

lina2011 [118]

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T = 24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

4 0

3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

GaryK [48]

Answer:

$F=4945.7N$

Explanation:

Most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces because these are the only forces on the horizontal direction the automobile is subjected to, besides of that generated by the engine itself. If the automobile is moving at constant speed, then these the force the engine provides is equal to the total friction force acting on the car.

We calculate then the force the engine provides. We know that power relates to work by the equation $P=\frac{W}{t}$ , and that work relates to force by the equation $W=Fd$ , so combining them we have:

$P=\frac{Fd}{t}=Fv$

Which can be written as:

$F=\frac{P}{v}$

And since $179hp$ are $179hp(\frac{746W}{1hp})=133534W$ , for our values the force the engine provides is:

$F=\frac{133534W}{27m/s}=4945.7N$

Which is also then the total friction force acting on the car.

6 0

3 years ago