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egoroff_w [7]
2 years ago
14

What determines the velocity of an object?

Physics
1 answer:
BARSIC [14]2 years ago
4 0
I think it’s A not 100% sure
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Which of the following is true about a planet orbiting a star in uniform circular
Mnenie [13.5K]

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.

7 0
3 years ago
Consider the accuracy of recording muscle electrical activity using electrodes placed on the skin surface, compared to directly
OverLord2011 [107]

Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method

Explanation:

Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:

--> Surface EMT and

--> Intramuscular EMT

Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.

Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.

The difference between the two forms of EMT includes:

- surface EMT is non- invasive while intramuscular EMT is invasive

- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.

- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.

8 0
3 years ago
The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

  • When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
  • While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
  • And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these  conclusions were made based upon the spot of glow on the fluorescent screen.

5 0
3 years ago
Read 2 more answers
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

5 0
3 years ago
SOMEONE PLEASE HELP ME ASAP PLEASE!!!! SIGNIFICANT DIGITS!!!!!! THE ANSWER IS NOT −16.422​
Aleks04 [339]

Explanation:

What exactly are u looking for?

7 0
3 years ago
Read 2 more answers
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