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2 years ago

A galaxy containing substantial amounts of dark matter will.

Physics

2 answers:

pogonyaev2 years ago

5 0

Answer:

A galaxy containing substantial amounts of dark matter will spin faster

AveGali [126]2 years ago

4 0

Answer:

spin faster

Explanation:

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A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?

anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}$

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

$\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}$

now, replace

$\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}$

hence, the answer is 0.495 m/s

3 0

1 year ago

What happens when a temperature <br>increases.

Liono4ka [1.6K]

The density decreases and convection causes the hot air particles to rise! BRANLIEST

5 0

3 years ago

Read 2 more answers

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

$t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s$

The force exerted by the floor on the ball can be found using the equation:

$F=\frac{\Delta p}{t}$

where

$\Delta p$ is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

$\Delta p = m(v-u)$

So the equation can be rewritten as

$F=\frac{m(v-u)}{t}$

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

$F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N$

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0

3 years ago

Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just s

tamaranim1 [39]

Answer:

Rolling Ball

Explanation

7 0

3 years ago

Space weather is dominated by a stream of charged particles emanating from the sun's magnetic field, known as ______________.

Katyanochek1 [597]

Answer:

<h2><em><u>solar wind</u></em></h2>

Explanation:

<em>This stream of particles emanating from sun's magnetic field is known as the solar wind. </em>

<em>The gas and particles come from the Sun's hot outer atmosphere, called the corona. These particles from the corona are charged with electricity.</em>

5 0

3 years ago