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3 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.

<u>Explanation:</u>

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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An iron ball weighs 80 N on the surface

Degger [83]

Answer:

2√5N will be the weight.

Explanation:

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2 years ago

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

6 0

2 years ago

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

3 years ago

During the first stage of sleep deprivation, the subject may __________.

klio [65]

<h2>First stage of sleep Deprivation Subject </h2>

During the first stage of sleep deprivation, the subject is NREM which stands for non-rapid eye movement. In this condition, we are not sleeping in the depth. It can be said as dreamless sleep. On electro-encephalography recording, the brain waves are not fast so they have high voltage.

In this condition, the breathing, heart rate and blood pressure is low. The sleep is comparatively tranquil. NREM lasts for 90 minutes to 120 minutes. It accounts for about 75% of the normal sleep time. Rapid eye movements do not occur.

7 0

2 years ago

Read 2 more answers

17. Which of the following is NOT a testable

Artemon [7]

Answer:

c. testing student opinions

Explanation:

opinions aren't factual and they would not aide an experiment if it wasn't for a social experiment.

7 0

2 years ago