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Roman55 [17]
3 years ago
5

How to measure the volume of a baseball bat ( need answers ASAP )

Physics
1 answer:
vaieri [72.5K]3 years ago
8 0

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

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9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en
masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

P_1 = Presión inicial = 0.96 atm

P_2 = Presión final

V_1 = Volumen inicial = 95 mL

V_2 = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0
2 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
Is a single property such as density sufficient to always correctly identify a substance?
bazaltina [42]
Not at all. Density on its own is not sufficient
7 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

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m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

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T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
How are primary and secondary succession similar and how are they different?
ExtremeBDS [4]
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8 0
3 years ago
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