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2 years ago

How many 200 ml cups can be filled with 6 1.5 litre bottles

1 answer:

6 0

What we need to know is that 1 liter is 1000 ml.

So, each liter can fill 1000/200=5 such cups.

We have how many liters?6*1.5=9

we have nine liters!

So, we will be able to fill 9*5=45

we will be able to fill 45 cups.

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The boy’s foot causes the motion. His foot is the one that causes the ball to roll down the hill.

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2 years ago

What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?

PIT_PIT [208]

We know the formula for density = Mass/ volume

So Mass, M = Volume * Density

Volume = 3.5 L= 0.0035 $m^3$

Density = 1.50 g/ml = 1500 $kg/m^3$

Mass, M = 0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

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3 years ago

Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ

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Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

$2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}$

$m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}$

= 221.79

= 221 (approx.)

hence no of bright fringe

M = m + 1

= 221 +1

M = 222 fringes

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2 years ago

What is voltage, and what is its relationship to amperage and power?

ziro4ka [17]

Explanation:

Amperage is the unit of electric current. It describes the strength of the electric current in a circuit.

The voltage is the driving force of the current in a circuit

Power is a function of voltage and current in the circuit.

Current is designate as I

Voltage as V

Power as P

I = $\frac{V}{R}$

Where R is the resistance to flow of electricity

P = I x V = $\frac{V^{2} }{R}$

The unit of power is watts and voltage is volts

learn more:

Voltage brainly.com/question/6949231

#learnwithBrainly

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The correct answer will be c

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