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2 years ago

Which of the following is true about a planet orbiting a star in uniform circular

1 answer:

7 0

The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.

A. The velocity vector of the planet points toward the center of the circle.

Explanation:

Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.

Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.

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When the compasses are moved close to a wire with a current which will the compasses point

Sindrei [870]

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2 years ago

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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2 years ago

Most chemical reactions can be divided into how many main groups?

tatyana61 [14]

There is synthesis
decomposition
double displacement
single displacement
combustion
metathesis
so i guess you could say 6

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2 years ago

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MrRa [10]

“Most climate scientists believe that there is evidence that explains global warming” is best supported by this study.

The drastic increase in the emission of CO2 (carbon dioxide) within the last 30 yearscaused by burning fossil fuels has been identified as the major reason for the change of temperature in the atmosphere (click the following link for a summary and graphs about the cause and effects of global warming).

The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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2 years ago

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A space shuttle is flying north at 6,510 m/s. It

Maksim231197 [3]

Answer:

7808 m/s

Explanation:

Find NE velocity after 60 s of acceleration in that direction:

= a t = 28.4 m/s^2 * 60 s = 1704 m/s

Vertical component = 1704 sin 45 = 1204.9 m/s

Horiz component = 1704 cos 45 = 1204.9 m/s

Add the two vertical components

6510 + 1204.9 = 7714.9 m/s = vertical velocity

Pythagorean theorem to find resultant of vertical and horiz v's

Vf ^2 = 1204.9^2 + 7714.9^2 0

Vf = 7808. m/s

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1 year ago