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3 years ago

How do galaxies vary?

1 answer:

4 0

The universe has trillions of galaxies and counting. Astronomers give names to each galaxy base on its shape (e.i, Sombrero galaxy, Milkyway Galazy, ect,.).

Also, the size of the galaxy is taken into account and their color.

You might be interested in

You are pushing a 20-kg box along a horizontal floor. Friction acts on the box. When you apply a horizontal force of magnitude 4

LenKa [72]

Answer:

$a = 2.6\,\frac{m}{s^{2}}$

Explanation:

The first box has the following equation of equilibrium:

$\Sigma F = F - f = 0$

$f = F$

$f = 48\,N$

The coefficient of friction is:

$\mu_{k} = \frac{f}{m \cdot g}$

$\mu_{k} = \frac{48\,N}{(20\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}$

$\mu_{k} = 0.245$

There is a net acceleration, when horizontal pushing force is increased:

$\Sigma F = F - f = m\cdot a$

$a = \frac{F-f}{m}$

$a = \frac{100\,N-48\,N}{20\,kg}$

$a = 2.6\,\frac{m}{s^{2}}$

7 0

3 years ago

Read 2 more answers

A wheel rotates with a constant angular acceleration of 3.45 rad/s^2. Assume the angular speed of the wheel is 1.85 rad/s at ti

lesantik [10]

Answer:

θ =607.33°

Explanation:

Given that

Angular acceleration α = 3.45 rad/s²

Initial angular speed ,ω = 1.85 rad/s

The angle rotates by wheel in time t

$\theta=\omega t +\dfrac{1}{2}\alpha t^2$

Now by putting the values

$\theta=\omega t +\dfrac{1}{2}\alpha t^2$

$\theta=1.85\times 2 +\dfrac{1}{2}\times 3.45\times 2^2$

θ = 10.6 rad

$\theta=\dfrac{180}{\pi}\times 10.6\ degree$

θ =607.33°

Therefore angle turn by wheel in 2 s is θ =607.33°

8 0

4 years ago

Suppose that you take a 10 kg mass on the surface of the earth and then place it on the moon. What will

sergey [27]

Answer:

10 kg

Explanation:

The mass of an object does not change even if the amount of gravtiy changes.

5 0

3 years ago

A body moves in circle cover half of revolution in 10 sec its linear distance became 10m along circumference of circle than its

stiks02 [169]

The centripetal acceleration is $0.63 m/s^2$

Explanation:

The centripetal acceleration for an object in circular motion is given by

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circle

The body in the problem cover half of revolution in

t = 10 s

And the corresponding linear distance covered is

L = 10 m

which corresponds to half of the circumference, so

$L=2\pi r$

From this equation we find the radius of the circle:

$r=\frac{L}{2\pi}=\frac{10}{2\pi}=1.59 m$ [m]

While the linear speed is:

$v=\frac{L}{t}=\frac{10}{10}=1 m/s$

Therefore, the centripetal acceleration is

$a=\frac{1^2}{1.59}=0.63 m/s^2$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

4 0

4 years ago

The Problems: 1. Xavier starts at a position of 0 m and moves with an average speed of 0.50 m/s for 3.0 seconds. He normally mov

NemiM [27]

Answer:

(1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

Explanation:

(1). Given that,

Initial position = 0 m

Average speed = 0.50 m/s

Time = 3.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Where, $x_{f}$ = final position

$x_{i}$ = Initial position

t = total time

Put the value into the formula

$0.50=\dfrac{x_{f}+0}{3.0}$

$x_{f}=0.50\times3.0$

$x_{f}=1.5\ m$

(2). Given that,

Initial position = 0 m

Average speed = 0.75 m/s

Time = 4.0 s

We need to calculate the final position

Using formula of average speed

$v_{av}=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$0.75=\dfrac{x_{f}+0}{4.0}$

$x_{f}=0.75\times4.0$

$x_{f}=3\ m$

(3). Given that,

Average speed = 1.25 m/s

Time = 3.0 sec

Initial position = 1.0 m

We need to calculate the final position

Using formula of average speed

$v=\dfrac{x_{f}+x_{i}}{t}$

Put the value into the formula

$1.25=\dfrac{x_{f}+1.0}{3.0}$

$x_{f}=1.25\times3.0-1.0$

$x_{f}=2.75\ m$

(4). Given that,

Average speed = 1.25 m/s

Distance = 100 m

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100}{1.25}$

$t=80 sec$

(5). Given that,

Average speed = 5 miles/hr

Suppose, distance = 25 miles

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25}{5}$

$t=5\ hr$

Hence, (1). His final position is 1.5 m.

(2). The final position of the hedgehog is 3 m.

(3). The final position of the tortoise

(4). Her race time is 80 sec.

(5). It take to finish in 5 hr.

5 0

4 years ago