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2 years ago

What occurs in the space when a massive object undergoes a change in its motion?

Physics

1 answer:

Lorico [155]2 years ago

7 0

Gravitational waves are emitted in the space when a massive object undergoes a change in its motion.

Gravitational waves are emitted. Bends spacetime, ripples travel outward from a gravitational source at the speed of light. The more massive the object and greater its acceleration, the stronger the resulting gravitational wave.

Therefore, A change in motion produces gravitational waves.

Definition of Gravitational waves:

Gravitational waves are disturbances or ripples in the curvature of spacetime, generated by accelerated masses, that propagate as waves outward from their source at the speed of light.

To learn more about Gravitational waves here

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A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to

shutvik [7]

k = 5.29

a = 0.78m/s²

KE = 0.0765J

<u>Explanation:</u>

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

4 0

3 years ago

Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.

jok3333 [9.3K]

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

= 9*3000

=27000ohm

I=VR

I=V/R

I= 120/27000

= 4.44*10⁻³A

4.44mA

Therefore, the current is 4.4mA

7 0

4 years ago

Which branch deals with anything foreign (other countries)<br> A executive b legislative c judicial

Schach [20]

It would be A. Executive branch

6 0

3 years ago

Letting D D represent the maximum displacement, the extremes of the block's motion are at position A, where x = − D x=−D, and at

Ksju [112]

Answer:

The answer is at x = 0, which represents position B

Explanation:

The full question is:

"A block is attached to a horizontal spring and set in a

simple harmonic motion, as shown from above in the figure. When the spring is relaxed, the block is a position B, where the displacement x from the equilibrium position is 0. Letting D represent the maximum displacement, the extremes of the block's motion are at position A, where x= -D, and at position C, where x= D.

At what point in the motion is the speed of the block at its maximum?"

And you can see the figure on the attached file.

Simple Harmonic motion equations

We can start from the equation that describes the position that is

$x(t)=D \sin\left(\omega t)$

Here D stands for the amplitude which is the maximum displacement, and $\omega$ is the angular velocity, thus we can find the derivative to find the velocity equation, so we get