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3 years ago

If the Moon rises at 7 A.M. on a particular day, then approximately what time will it rise four days later?

Physics

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

10:33 am

Explanation:

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The set of frequencies of the electromagnetic waves emitted by the atoms of an element is called

VLD [36.1K]

Answer:

The set of frequencies of the electromagnetic Waves emitted by the atoms of an element is called emission spectrum.

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3 years ago

A 241 kg mass is lifted 1.8 m. What is the potential energy of the mass (in J)?

Korolek [52]

Answer:

mass is lifted 1.8 m. What is the potential energy of the mass 4. A 100 kg

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3 years ago

Read 2 more answers

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

$P = I^2R$

make I subject of the formula we have

$I= \sqrt{\frac{P}{R} }$

substitute

$I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps$

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3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

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3 years ago

Gumuhit ng paglalaruan ng<br>agawang sulok at ibigay ang bawat<br>sukat nito.

Anarel [89]

Answer:

there's a link to answer

Explanation:

there's a link to answer in another's persons response it helps a lot just click on it

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3 years ago