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Mama L [17]
2 years ago
6

You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

aws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of g, the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height h of the top of the incline above the lower end of the incline can be varied. You measure the speed vcm of the center of mass of the cylinder when it reaches the bottom for various values of h. You plot v2cm (in m2/s2) versus h (in m) and find that your data lie close to a straight line with a slope of 6.42 m/s2
Physics
1 answer:
Ugo [173]2 years ago
3 0

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

E_p = E_{kv} + E_{k\omega}

mgh = mv^2/2 + I\omega^2/2

where m is the mass, I = mr^2/2 is the moments of inertia of the solid cylinder \omega = v / r is the angular speed of the cylinder

mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}

mgh = mv^2/2 + mv^2/4 = 3mv^2/4

h = 3v^2/(4g)

So if you plot a liner chart of h vs v^2 and get a slope of 6.42 then that means 3/(4g) = 6.42 so g = 6.42*4/3 = 8.56 m/s^2

The gravitational acceleration on this planet is 8.56 m/s2

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Rus_ich [418]
M1 descending
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m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

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   = (5.60 − 2)/(2 + 5.60) x 9.81 
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5 0
2 years ago
What principles of science (like facts, laws, and theories) might help explain why similar investigations conducted in many part
natulia [17]

Answer:

Reproducibility of research

Explanation:

The principle of science that explains why similar experimental investigations conducted in different parts of the world could result in the same outcome is referred to as reproducibility.

<em>A good research or experiment in science must be reproducible, otherwise, the outcome of such an experiment might become inadmissible within the scientific community. It is a core principle of the scientific method that similar results should be obtained when an experiment or observational study conducted in one place is repeated in another place with the same procedure. Hence, an experiment must be reproducible in science in order for the outcome of such an experiment to be part of the general scientific knowledge. </em>

7 0
1 year ago
Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio
leva [86]

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B ) Ascend using my buddy alternative air source / make an emergency Ascent

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5 0
2 years ago
Read 2 more answers
1. A particular lever is 90.0% efficient. If 50.0 J of work are done on the lever, then how much work does the lever do on its l
laila [671]

Answer:

Explanation:

Using the efficiency formula;

Efficiency = Work done by the machine (output)/work done on the machine (input) ×100%

Efficiency =w/50 ×100

90 = 100w/50

Cross multiply

90×50 = 100W

4500 = 100W

W = 4500/100

W = 45Joules

Hence the lever does 45Joules of work on its load

2) Mechanical Advantage= Load/Effort

Given

MA = 4

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4 = 500/Effort

Effort = 500/4

Effort =125N

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8 0
1 year ago
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz
Olin [163]

Answer:

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According to Doppler's effect of sound we have

for a source of sound moving away from the observer the relation between the observed and the original frequency is given by

f_{app}=\frac{c-v_{rec}}{c+v_{s}}\times f_{original}

where

c = speed of sound in air

v_{rec} is the velocity of observer of sound

v_{s} is the velocity of source of sound

f_{o} is the original frequency of sound

As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.

2) <u>Effect on Intensity:</u>

At a distance 'r' from source emitting a wave of Power 'P' is given by

I=\frac{P}{4\pi r^{2}}

As we see on increasing 'r' intensity of sound decreases.

3 0
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