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Brilliant_brown [7]

3 years ago

8

Do any of the galaxies appear to be closer to each other as your universe expands? What does this say about the possibility of g

alaxies colliding with each other?

1 answer:

Luda [366]3 years ago

8 0

Explanation:

In local galactic group the force of expansion of universe is overcome by the force of attraction due to gravity. Best example is our own galaxy milky way and another giant galaxy in our local group Andromeda. Andromeda having enormous gravity is pulling milky way towards itself, overcoming the force of expansion.

So, there are possibilities of collision despite the expansion of universe at a rapid pace. It is estimated that the milky way and Andromeda will collide each other after about 50 billion years from now.

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What information and measurements would you need to calculate the rate of movement?

aliina [53]

When you talk about rate, you will expect that it will be in terms of a time unit. It measures how fast it is going. So, you would expect that the denominator is in time units. For the movement, you can measure this with either distance or velocity.

So, for the first variety, you would need distance and time to measure the rate of how far you go at a certain time. It is also called as velocity. For the second variety, you would need velocity and time to measure the rate of how fast you are going at a certain interval. It is also called as acceleration.

8 0

3 years ago

gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the

igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

$y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\$

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

$d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\$

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

7 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

What is the momentum of a two-particle system

Harlamova29_29 [7]

Answer:

-67,500 kgm/s

Explanation:

1300 * 20 + 1100 * (-85) = -67,500 kgm/s

8 0

3 years ago

A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel

7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

$\omega = \frac{\theta}{t}$

Here,

$\theta$ = Angular displacement

t = time

The angular velocity is given as

$\omega = 230rev/min$

PART A) The angular velocity in SI Units will be,

$\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s$

PART B) From our first equation we can rearrange to find the angular displacement then

$\theta = \omega t$

Replacing,

$\theta = (24.08)(0.75)$

$\theta = 18.06 rad$

4 0

3 years ago