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maria [59]
3 years ago
11

Riddle of the day What gets wet while drying?

Physics
2 answers:
Ulleksa [173]3 years ago
4 0

Answer:

A towel!

Explanation:

The towel gets wet when it dries something off!

garri49 [273]3 years ago
4 0

Answer:

lol a towel?

Explanation:

because like- I don't know but you get where I'm coming from right?

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What is power?
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Answer:

B) Power is the rate at which work is done

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4 years ago
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I need help ASAP. This is for 15 points
kramer

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

7 0
3 years ago
If you hold a 50 kilogram barbell above your head for 3 seconds and
aalyn [17]

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c

if you calculate the net force you get 490 N

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3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
3 years ago
A 120 V fish-tank heater is rated at 130W. Calculate (a) the current through the heater when it is operating, and (b) its resist
7nadin3 [17]

Explanation:

The power P dissipated by a heater is defined as

P = VI

where V is the voltage and I is the current.

a) The current running through a 130-W heater is

I = \dfrac{P}{V} = \dfrac{130\:\text{W}}{120\:\text{V}} = 1.08\:\text{A}

b) The resistance <em>R</em><em> </em>of the heater is

P = VI = (IR)I = I^2R

where V= IR is our familiar Ohm's Law.

\Rightarrow R = \dfrac{P}{I^2} = \dfrac{130\:\text{W}}{(1.08\:\text{A})^2}

R = 110.8\:Ω

8 0
3 years ago
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