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Elena-2011 [213]

3 years ago

8

A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0

m/s . what is the acceleration
A 3.3 m/s^2
B 26.6 m/s^2
C 13.3 m/s^2
D 6.6 m/s^2

1 answer:

OLEGan [10]3 years ago

7 0

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

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A stone is dropped from rest from the top of a building. It takes Î”t = 2.2 s for it to reach the ground.

NeTakaya

Answer:

Value of magnitude of acceleration will be $9.8m/sec^2$

Explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity $u_i$ will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to $9.8m/sec^2$

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3 years ago

Please help- it’s a 15 point change to my grade.

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Answer: Everything except heat and density

Explanation:

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2 years ago

A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

inna [77]

This is a defective, misleading question, and should never be asked in a Physics class.

There is no such thing as the force due to the impact.

If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.

It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.

I would kick this question back to the Physics teacher. It's meaningless,
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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

4 0

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juin [17]

I think the answer is c but I’m not sure

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2 years ago