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Dennis_Churaev [7]
3 years ago
14

A 144-g baseball is dropped from a tree 11.0m above the ground.A.With what speed would it hit the ground if air resistance could

be ignored?Express your answer to three significant figures and include the appropriate units.B.If it actually hits the ground with a speed of 7.50m/s , what is the magnitude of the average force of air resistance exerted on it?Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
nikitadnepr [17]3 years ago
5 0

Answer:

a. 14.68m/s

b. 1.043N

Explanation:

a.

Given

Mass = m = 144g = 0.144kg

Height = h = 11m

Using energy conservation,

½mv² = mgh where g = 9.8m/s² ----- divide both sides by m

½v² = gh ---- divide both sides by ½

v² = 2gh

v = √2gh

v = √(2 * 9.8 * 11)

v = 14.68332387438212m/s

v = 14.68m/s ----- Approximated

b.

v = 7.5m/s

By definition, “the average force of air resistance,”

When work done against air resistance is solved by multiplying height by the average force of resistance ------ by definition

i.e. W = -Fh

Minus is used because F and h points to different direction (when viewed as vectors).

This is proved by:

W = Fhcos(180) because h points toward the direction of motion (downward) while F points opposite.

W = Fh * -1

W = Fh

Using energy conservation,

mgh - Fh = ½mv² ---- Make Fh, the subject of formula

Fh = mgh - ½mv² ---- Divide through by h

F = mg - mv²/2h

F = 0.144 * 9.8 - 0.144 * 7.5²/(2 * 11)

F = 1.043N

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Explanation:

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In both cases plates are moving away from one another. Therefore they are creating new land masses.

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3 years ago
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Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa
Lorico [155]

Answer:r_2=11.81 cm

Explanation:

Given

m_1=2.2\times 10^{-8} kg

m_2=4.8\times 10^{-8} kg

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potential Energy due to Magnetic Field =Kinetic Energy of Particle

qV=\frac{mv^2}{2}

v=\sqrt{\frac{2qV}{m}}

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Force due to magnetic field will Provide centripetal Force

qvB=\frac{mv^2}{r}

B=\frac{\sqrt{\frac{2Vm}{q}}}{r}

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thus \frac{m}{r^2}=constant

\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}

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A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
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Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

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Answer:

d. 50 C

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According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.

The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:

Final temperature = (20 C + 80 C)/2

= 50 Celsius

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3 years ago
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