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2 years ago

14

Lindsay is boiling macaroni noodles in a pot of water. The noodles rise and fall as the thermal energy currents move from areas

of higher energy to areas of lower energy. Which process is primarily responsible for making the macaroni noodles move in the boiling water?

1 answer:

k0ka [10]2 years ago

4 0

The process primarily responsible for noodle motion, especially in this scenario, is convection.

In any fluid undergoing uneven heating or cooling, a circulation pattern develops where warmer portions of the fluid rise and cooler portions sink.

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In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3

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Answer:

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You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

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The speed of the truck is $v = 22.6\ m/s$

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$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

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=> $\Delta KE = 0.5 * m * 22.6^2$

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2 years ago

A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must

mezya [45]

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So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)

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2 years ago

Read 2 more answers

Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V

Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

$Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter$

$Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter$

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

$R=\dfrac{\rho\times l}{A}$

Where,

R = Resistance

l = length

A = Area of cross section = πr²

$\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter$

Substituting the values we get

$R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}$

$R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}$

$R=\dfrac{1\times 10^{2}}{3.23}$

$R=30.95\ ohm$

Now by Ohm's Law,

$V= I\times R$

Substituting the values we get

$I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere$

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0

2 years ago