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3 years ago

You walk out to the edge of a 2,384 m. cliff and you toss a rock down to watch it drop. What is the rocks acceleration?

Physics

1 answer:

likoan [24]3 years ago

3 0

Answer:

9.8m/s

Explanation:

Wherever you're on the earth's surface, gravitational force remains the same and it does not matter either its a rock that's being tossed or a sheet of paper, the acceleration remains 9.8m/s only. *Condition applied=air resistance neglected

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If all the light waves are reflected in equal amounts off an object then the object will appear

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The object would appear bright

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A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is

borishaifa [10]

Answer:

6.4m/s

Explanation:

The total mechanical energy of the man is 1780J.

This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.

Also, mass = 87kg

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

1780 = $\frac{1}{2}$ x 87 x v²

v² = 40.9

v = 6.4m/s

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3 years ago

A typical cell has a membrane potential of -70 , meaning that the potential inside the cell is 70 less than the potential outsid

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Answer:

energy is stored is 2.2 × 10⁻¹³ J

Explanation:

The capacitance of the cell is given with the expression

C = (KE₀A) / d

k is the dielectric constant, A is the area of the cell, d is the thickness of the cell.

Now given that; the diameter is 50,

Area A = 4πR²

A = 4π × ( 25 × 10⁻⁶ m)²

A = 7850×10⁻¹² m²

our capacitance C = (KE₀A) / d

C = [9 ( 8.85 × 10⁻¹² C²/N.m² × 7850×10⁻¹² m² )] / 7×10⁻⁹ m

C = 8.93 × 10⁻¹¹ F

Now Energy stored

E = 1/2 × CV²

E = 1/2 × (8.93 × 10⁻¹¹ F) × ( 70 × 10⁻³ V)²

E = 2.2 × 10⁻¹³ J

Therefore energy is stored is 2.2 × 10⁻¹³ J

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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo

faltersainse [42]

Answer:

Length of pipe $= 0.057$ meter

Explanation:

Speed of a transverse wave on a string

$v = \sqrt{\frac{F}{\mu} }$

where F is the tension in string and $\mu$ is the mass per unit length

Thus,

$\mu = \frac{m}{L}$

Substituting the given values we get -

$\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}$

Speed of a transverse wave on a string

$v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}$

For third harmonic wave , frequency is equal to

$f = \frac{nv}{2L}$

Substituting the given values, we get -

$f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08$

Length of pipe

$L = \frac{nv}{4 f}$

Substituting the given values we get

$n = 1$ for first harmonic wave

$L = \frac{344* 1}{4*1502.08} \\L = 0.057$

Length of pipe $= 0.057$ meter

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3 years ago