The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?

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Answer 1

Answer:

The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems

Related Questions

24-gauge copper wire has a diameter of 0.51 mm. The speaker is located exactly 4.27 m away from the amplifier. What is the minimum resistance of the connecting speaker wire at 20°C? Hint: How many wires are required to connect a speaker!Compare the resistance of the wire to the resistance of the speaker (Rsp = 8 capital omega)
A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables
Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?
Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

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The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

$v = \omega r$ ,

Where r indicates the radius and $\omega$ the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

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Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

A car is decelerating at the rate of 2 km/s square. If its initial speed is 66 km/s, how long will it take the car to come to a complete stop?

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Answer:

It will take 33 seconds to stop the car.

Explanation:

Using the first equation of kinematics we have

$v=u+at$

where

'v' is final speed of object

'u' is initial speed of object

'a' is acceleration of object

't' is time of acceleration of object

Now since it is given that $a=-2km/s^(2)$ since acceleration is negative and $u=66km/s$

We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get

$0=66-2* t\n\n\therefore t=(66)/(2)=33seconds$

A firefighting crew uses a water cannon that shoots water at 27.0 m/s at a fixed angle of 50.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1Part A:
d1=_____m
Part B:
d2=______m

Answers

Answer:

Explanation:

In projectile motion , range of projectile is given by the expressions

R = u²sin2θ / g

where u is velocity of projectile.

u = 27 m/s θ = 50

12 = 27² sin 2θ / 9.8

sin 2θ = .16

θ = 9.2 / 2

= 4.6

When we place 90- θ in place of θ , in the formula of range , we get the same value of projectile. hence at 85.4 ° , the range will be same.

Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?