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Citrus2011 [14]

2 years ago

5

In the first direct detection of gravitational waves by LIGO in 2015, the waves came from A. the collapse of a nearby star into

a white dwarf B. the merger of two black holes C. a supernova explosion in a nearby galaxy D. the rapid motion of three hot Jupiter planets around a

1 answer:

diamong [38]2 years ago

8 0

In the first direct detection of gravitational waves by LIGO in 2015, the waves came from the merger of two black holes. Option B is correct. This is further explained below.

<h3>What are gravitational waves?</h3>

A gravitational wave is simply defined as a ripple in space that is unseen though extremely rapid. Gravitational waves move at light speed. As they pass past, these waves compress and stretch everything in their path.

In conclusion, the merger of two black holes is the first direct detection of gravitational waves.

Read more about Wave

brainly.com/question/23271222

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A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W

Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s$

Since, the motion is downward, final velocity must be negative. So,

$v_d=-4.95\ m/s$

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity ( $v_d$ ) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

$v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s$

Since, the motion is upward, final velocity must be positive. So,

$v_{up}=3.31\ m/s$

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

$J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns$

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

5 0

2 years ago

1) Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a

masha68 [24]

Answer:

Could you explain that more better?

Explanation:

3 0

3 years ago

If the police are chasing a runaway car why is knowing the velocity of the car important

atroni [7]

If you only know its speed, that's not enough information to catch it. You could even chase it at DOUBLE that speed, and you'd never catch it if you were chasing in the wrong direction.

You also have to know the DIRECTION the runaway car is going, so that you can chase in the same direction.

Now that you know its speed AND direction, you know its velocity. You need that information to have any chance of catching it.

8 0

2 years ago

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same dire

Dahasolnce [82]

Answer:

v₁ =0.19 m/s and v₂ = 0.18 m/s

Explanation:

By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

<u>Momentum:</u>

Before (b) = After (a)

$m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}$

$26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ $7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ (1)

<u>Energy:</u>

Before (b) = After (a)

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$

$26.5 kg*(0.205 m/s)^{2} + 12.5 kg*(0.150 m/s)^{2} = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$ (2)

<u>From equation (1) we have:</u>

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}$ (3)

<u>Now, by entering equation (3) into (2) we have: </u>

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}$ (4)

By solving equation (4) for $v_{1a}$ , we will have two values for

$v_{1a} = 0.16$

$v_{1a} = 0.21$

We will take the average of both values:

$v_{1a} = 0.19 m/s$

Now, by introducing this value into equation (3) we can find $v_{2a}$ :

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}$

$v_{2a} = 0.18 m/s$

Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

I hope it helps you!

6 0

2 years ago

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

stepladder [879]

Answer:

Capacitor, is the right answer.

Explanation:

The unknown element is a Capacitor.

Below is the calculation that proves that it is a capacitor.

We know that for the Capacitor

i = Imax × sin(wt+(pi/2)).

i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))

i = Imax × sin(3.142) = 0 A

at, t = T/2

wt = (2 × pi/T) × (T/2) = pi

wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =

i = Imax × sin(3 × pi/2) = -Imax

Which is in a correct agreement with capacitor therefore, the answer is a Capacitor.

5 0

2 years ago