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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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Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

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r_1 = 0.6875 k ohm

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}

r_2 = 0.95 k ohm

now the current flowing through the battery is

i = \frac{2}{1 + 0.95} = 1.02 mA

now this will divide into R3 and R2 so current flowing in R3 will be

i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA

now this will again divide in R4 and R5

so current in R5 will be

i_5 = \frac{R_4}{R_4 + R_5}* i_1

i_5 = 0.18 mA

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm

also after its series with R3

r_2 = 1 + 0.6875 = 1.6875 k ohm

now it is in parallel with R5 on other side

r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm

now current through the battery will be given as

i = \frac{3}{1 + 0.95} = 1.53 mA

now it is divide in r2 and R5

so current in R5 is given as

i_5 = \frac{r_2}{r_2 + R_5}*i

i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53

i_5 = 0.67 mA

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i = 0.67 + 0.18 = 0.85 mA

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2 years ago
Describe a well-known hypothesis that was discarded because it was found to be untrue.earth-centered model of the universe. the
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The widely accepted hypothesis before that turned out wrong was the Earth-Centered theory or the Geocentric Theory. This was proposed by the philosopher Ptolemy. He came about to this hypothesis from hi observation that from the Earth's perspective, the celestial bodies like the Sun, stars and the moon, look like they rotate around the Earth each day and night. However, this was disproved by Galileo Galelei by his Heliocentric Theory. He observed through the telescope that the Venus also changes phases like the moon. However, he deduced that this is not possible from the positions of the Venus, Earth, Moon and Sun. 
6 0
2 years ago
One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned
liq [111]

Answer:

T = 3183 N

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

\Delta L = 2(2\pi r)

\Delta L = 4\pi r

\Delta L = 4(\pi)(1.0 mm)

\Delta L = 12.56 mm

now we know by the formula of Young's modulus

Y = \frac{T/A}{\Delta L/L}

so we have

T = \frac{AY \Delta L}{L}

T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}

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2 years ago
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi
KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, A=7.85\times 10^{-5}\ m^2

Resistivity of copper, \rho=1.72\times 10^{-8}\ \Omega-m

The resistance of a wire is given by :

R=\rho\dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}

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Answer:

Explanation:

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d = 25.66 cm

5 0
2 years ago
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