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2 years ago

Please answer please please Physics

Physics

1 answer:

zmey [24]2 years ago

6 0

Answer:

about -1.26 m/s²

Explanation:

The acceleration can be found from the formula ...

a = (v2² -v1²)/(2d)

Here, we have the initial velocity v1=23 m/s, the final velocity v2=0, and teh distance d=210 m. Using the formula, we find the acceleration to be ...

a = (0 -23²)/2(210) ≈ -1.2595 . . . m/s²

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A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

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v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

3 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

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3 years ago

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timofeeve [1]

Answer:

A microwave

Explanation:

Car

Lightbulb is a really good one

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3 0

1 year ago

Explain the relationship between the current output of the power supply and the current through each component in the parallel c

baherus [9]

Explanation:

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