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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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The resistance of a wire depends on

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Answer:

Explanation:

because all are factors that determine the resistivity of a material

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3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

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