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nika2105 [10]
3 years ago
6

A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o

f mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
Physics
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

-use the exact same equipment

-use the exact same measurements (dropping height)

Explanation:

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The nearest star to our planet other than the sun , is 4.4 light year away.One light-year is the distance light travels in a yea
lozanna [386]

Answer:

1606

The main reason for using light years, however, is because the distances we deal with in space are immense. If we stick to miles or kilometers we quickly run into unwieldy numbers just measuring the distance to the nearest star: a dim red dwarf called Proxima Centauri that sits a mere 24,000,000,000,000 miles away!

Explanation:

3 0
3 years ago
Can someone help me?!!!!!
german
<h2>Hello!</h2>

The answer is:

The first option,  the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

Distance=NorthCoveredDistance+EastCoveredDistance

Distance=4*180m+3*180m=720m+540m=1260m

Actual distance between the start and the end point (displacement):

ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0
3 years ago
1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and
saul85 [17]

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

   Centripetal acceleration  = \frac{v^{2} }{r}

   v is the velocity of the body

    r is the radius of the circle

  putting in the parameters:

   Centripetal acceleration = \frac{4^{2} }{0.75}

    Centripetal acceleration = 10.67m/s²

Centripetal force = m  \frac{v^{2} }{r}

          m is the mass

 Centripetal force = mass x centripetal acceleration

                              = 3 x 10.67

                              = 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

4 0
3 years ago
What is a molecule called when it has more than one element?
Alinara [238K]
The correct answer is compound
4 0
3 years ago
Read 2 more answers
Assignment<br> S<br> of<br> write the Symbol<br> Told, mercury and<br> Cooper, Iron,<br> Lead
m_a_m_a [10]

Answer:

Check explanation

Explanation:

Gold - Au (Aurum)

Mercury - Hg (Hydrargyrum)

Copper - Cu (Cuprum)

Iron - Fe (Ferrum)

Lead - Pb (Plumbum)

These elements in the periodic table are some of the elements represented by letters not in line with their names.

This is because, these elements were known in ancient times and therefore, they are represented by letters from their ancient names.

3 0
3 years ago
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