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1 year ago

The Andromeda Galaxy (our nearest spiral neighbor) has spectral lines that show a blue shift. From this we may conclude that:

1 answer:

8 0

The universe is the collection of galaxies, The Andromeda Galaxy has spectral lines with blue shift. The conclusion is that the Universe has stopped expanding.

<h3>What is Andromeda Galaxy?</h3>

The Andromeda Galaxy is the nearest spiral neighbor that has spectral lines showing a blue shift.

Therefore, this concludes that, the Universe has stopped expanding. This galaxy is slowly shifting towards us.

Learn more about Andromeda Galaxy.

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Coal burns in a furnace, producing light and heat. This reaction is

olasank [31]

The reaction is Exothermic

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2 years ago

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A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw

irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0

3 years ago

Fa car's power output is increased, its efficiency:

Inessa [10]

It’s solved by using a pretty standard formula for efficiency.

4 0

2 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

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3 years ago

A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal

SpyIntel [72]

Answer:

$x=1.75m$

Explanation:

From the exercise we have that

$y_{o}=1.3m\\v_{o}=3m/s, \beta =37\\$

<em><u>To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first </u></em>

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0m

$0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=-0.36 s$ or $t=0.73s$

Since the <u>time</u> can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

$x=v_{ox}t=3cos(37)(0.73)=1.75m$

6 0

3 years ago