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3 years ago

Compared to the atoms in a room-temperature banana, the atoms in a frozen banana have _____ kinetic energy.

Physics

1 answer:

Tju [1.3M]3 years ago

3 0

The answer is b. more

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According to the first rule, if a force pulls on one end of a rope, the tension in the rope equals the magnitude of the pulling

mixer [17]

Answer:

Explanation:

Here in the question the mass of the pulley is zero, hence, the tension in the cable throughout is same.

magnitude of tension in rope 1 is

T1= F

Hence the tension T1 is rope 1 is F.

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3 years ago

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What can electromagnetic waves travel through?

Kisachek [45]

Electromagnetic waves do not require a medium to travel through. They can travel through empty space or matter.

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2 years ago

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap

Verizon [17]

Answer:

The answer is " $1.144 \times 10^{-4} \ m$ ".

Explanation:

$w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\$

$=\frac{2.44 \times 633 \times 10^{-9}\times 4 }{0.054}\\\\=\frac{6178.08\times 10^{-9}}{0.054}\\\\=1.144 \times 10^{-4} \ m$

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3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

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2 years ago

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Which environmental change occurs quickly

never [62]

Air and water have a good day

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2 years ago