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3 years ago

Which region of the early universe was most likely to become a galaxy?

Physics

1 answer:

kramer3 years ago

3 0

Answer:

This is likely possible for a region whose matter density is higher than the normal average.

Explanation:

A galaxy is a collection of lumps in space which are clumped together and interact with each other. There are a lot of speculations on how galaxies were birthed. some believe its formed by a collection of massive gas, dust which eventually collapsed under their own gravitational pull. others says its formed by the combination of large lumps of matter which accumulated forming thee galaxies. The possibility of a galaxy forming is dependent on how massive the matter in the region of the universe is.

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75 kilometers per hour is an example of:<br> A. speed.<br> B. velocity<br> C. acceleration

kherson [118]

The answer is A. speed
hope this helps! :D

5 0

3 years ago

Read 2 more answers

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass

Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown:

Velocity of mass = ?

Kinetic energy = ?

Solution:

Velocity is the rate of change of displacement with time. It is a vector quantity that shows magnitude and direction.

Mathematically;

Velocity = $\frac{displacement}{time}$

Velocity = $\frac{5}{5.8 }$ = 0.86m/s

Kinetic energy is the energy due to the motion of a body. It is expressed mathematically as:

Kinetic energy = $\frac{1}{2} m v^{2}$

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2 } x 0.343 x 0.86^{2}$ = 0.13J

learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

7 0

3 years ago

Assume that two cars have the same kinetic energy, but that the red car has twice the speed of the blue car. We then know that t

Anna71 [15]

Answer:

equal

Explanation:

8 0

2 years ago