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3 years ago

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If Michael Jordan has a vertical leap of 1.29 meters, what is his take-off speed and his hang time (total time to move upwards t

o the peak and then return to the ground)?

1 answer:

evablogger [386]3 years ago

6 0

(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>

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Can a conductor be given limitless charge

k0ka [10]

Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

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2 years ago

Struggling on this, really need help

Bad White [126]

Answer:

Guessing you just need help with the definition but if it's the question I can still help you.

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2 years ago

Read 2 more answers

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

2 years ago

Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is

goblinko [34]

Answer:

d $\frac{x}{l}$ = m× λ⇒ d = λ ×m×l / x

= 630× $10^{-9}$ m × 3×3m/ 45× $10^{-3}$ m

= 1.26× $10^{-4}$ m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

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2 years ago

Please could someone explain this.

icang [17]

This problem is to let you practice using Newton's second law of motion:

Force = (mass) x (acceleration)

-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.

-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.

-- Now you know ' F ' and ' mass '. Use Newton's second law of motion to calculate the plane's acceleration.

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3 years ago