4 years ago

What is an ore? an object that has been recycled

Physics

1 answer:

motikmotik4 years ago

6 0

an ore is a naturally occurring solid material from which a metal or valuable mineral can be profitably extracted.

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A current of 0.92 A flows past a point in a circuit? How much charge, in units of Coulombs, passes that point in one minute?

Alisiya [41]

<h3><u>Answer;</u></h3>

<u> = 55.2 Coulombs </u>

<h3><u>Explanation</u>;</h3>

We can determine Charge using the formula

Q =It, where Q is the amount of charge in Coulombs, I is the current in amperes and t is the time in seconds.

I = 0.92 amperes, t = 1 minute or 60 seconds

Charge = 0.92 × 60

<u> = 55.2 Coulombs </u>

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3 years ago

A _______ contains several blades of differing thicknesses and is used to measure the clearance between parts. A. dial in

Likurg_2 [28]

This is D. Feeler Gage

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3 years ago

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v

ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt) = 0.5

$\omega t = sin^{-1}(0.5)$

$\dfrac{2\pi}{T}\times t =0.524$

$\dfrac{2\pi}{1.5}\times t =0.524$

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

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3 years ago

How do longitudinal waves transport energy?

marissa [1.9K]

A longitudinal wave transports energy through the medium without permanently transporting matter.

Hope this helps :D

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3 years ago

What is the centripetal acceleration of a small laboratory centrifuge in which the tip of the test tube is moving at 19.0 meters

Artist 52 [7]

Answer: $3.61(10)^{3} \frac{m}{s^{2}}$

The centripetal acceleration $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$

Where:

$V=19m/s$ is the velocity

$r=10cm=0.1m$ is the radius of the circle

$a_{c}=\frac{(19m/s)^{2}}{0.1m}$

$a_{c}=3610m/s^{2}=3.61(10)^{3}m/s^{2}$

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3 years ago