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1 year ago

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

Physics

1 answer:

Lorico [155]1 year 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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HELP the decay curve of twizzlers sheet!!!!!

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Ok so the first thing you need to do is im not gonna help you

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3 years ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

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3 years ago

Which of the following is the number of cycles per unit of time? infrasonis wave,frequency,ultrasonic wave or pitch

Allushta [10]

Frequency
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3 years ago

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A spinning ice skater will speed up if he brings his arms close to his body. Which of the following statements explains this phe

xxMikexx [17]

A. Angular momentum is always conserved would be the correct answer.

This is because like linear momentum (mvmv), angular momentum (r×mvr×mv) is a conserved quantity, where rr is the vector from the center of rotation. For a skater holding a static pose, for each particle making up her body, the contribution in magnitude to the total angular momentum is given by mirivimirivi. Thus bringing in her arms reduces riri for those particles. In order to conserve angular momentum, there is then an increase in the angular velocity.

hope this helps!

7 0

3 years ago

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