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2 years ago

A galaxy with a lot of dark matter would have a high mass-to-light ratio compared to the Sun. True False

Physics

1 answer:

Ivenika [448]2 years ago

8 0

Yes, a galaxy with a lot of dark matter would have a high mass-to-light ratio compared to the Sun.

<h3>Contribution of dark matter in mass to light ratio?</h3>

This largely invisible matter such as dark matter adds to the mass of the galaxy while it has no contribution in its luminosity which increases the mass-to-light ratio. If dark invisible matter is present in a galaxy, its mass-to-light ratio can be as high.

So we can conclude that the statement is true.

Learn more about galaxy here: brainly.com/question/13956361

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what is the kinetic energy of an object that as a mass of 12 kilograms and moves with the velocity of 10 m/s

Galina-37 [17]

Answer:

120 m/s

Explanation: QUIZLET

7 0

3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

Convert the following as instructed:

Elis [28]

3.4m
6700g
172800s
1 day

7 0

3 years ago

The resistance of resistor is greater for:

Monica [59]

Answer:

c: long and thin resistor.

Explanation:

The resistance of a resistor is given by:

R = ρ*L/A

where:

R = resistance

ρ = resistivity (depends on the material)

L = length of the material

A = cross-sectional area of the material

We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.

We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.

Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.

8 0

3 years ago

Can someone explain to how to calculate this

Karo-lina-s [1.5K]

answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

3 0

3 years ago