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2 years ago

6

Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t

he nozzle of the hose if it has an Inside dlameter of 6.0 mm?

1 answer:

Leya [2.2K]2 years ago

3 0

Answer:

15 m/s

Explanation:

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PLS HELP IM SOO CONFUSED!

Feliz [49]

False.

As temperature increases the more the electrons begin to vibrate more, as it decreases they vibrate less.

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2 years ago

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A pulley has a mechanical advantage of 1.

sertanlavr [38]

Answer:

There is no mechanical advantage

Explanation:

The mechanical advantage is possible only when the force needed to lift a load is lesser than the weight of the load.

For example, is we have a mechanical advantage of 2, the force needed to lift will be 1/2 of the weight of the load, and if we have a mechanical advantage of 4, the force needed will be 1/4 of the weight of the load.

In the attached image there are clear examples of mechanical advantage with pulleys.

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2 years ago

What is the weight of a 4.2 kg bowling ball on Mars?

Nataliya [291]

What is the weight of a 4.2 kg bowling ball on Mars?

Answer:

1.59 kg

Explanation:

The formula is:

<u>F = G((Mm)/r2) </u>

F is the gravitational force between two objects,

G is the Gravitational Constant (6.674×10-11 Newtons x meters2 / kilograms2),

M is the planet's mass (kg),

m is your mass (kg), and

r is the distance (m) between the centers of the two masses (the planet's radius).

Hope this helps

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3 years ago

Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks

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Answer: The same

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At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

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3 years ago