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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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The body went 450 meters within 30 seconds of starting the movement. In what time did the first 50 meters go?

Andrew [12]

3.33 seconds.

<u>Explanation:</u>

We can find the speed of the body using the formula,

Speed = Distance traveled in meters / time taken in seconds

= 450 m / 30 seconds

= 15 m/s

So per second the distance traveled by the body is 15 m.

So time needed to travel 50 m can be found as,

time = distance/speed

= 50 m / 15 m /s

= 3.33 s

8 0

2 years ago

This is what occurs when matter transitions between solid, liquid and gas.

vodomira [7]

Answer:

The answer is Phase Change

Explanation:

4 0

3 years ago

Read 2 more answers

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

$\epsilon = BvL sin \theta$

where

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

$\theta=40^{\circ}$ is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

$\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV$

8 0

2 years ago

An object undergoes two successive displacements:

Gre4nikov [31]

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

$d_1=79 m$ at $16.9^{\circ}$

So its components are

$d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m$

2nd displacement is:

$d_2=16.7 m$ at $31.1^{\circ}$

So its components are

$d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m$

Therefore, the x- and y-components of the net displacement are:

$d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m$

Therefore, the magnitude of the final displacement is:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m$

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

8 0

2 years ago

Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive or attractive only

Y: Very small range

Z: Repulsive and attractive

8 0

3 years ago