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sweet [91]
3 years ago
8

if a big stone and a small stone is dropped simultaneously from the certain height. they fall together why??​

Physics
2 answers:
Fiesta28 [93]3 years ago
5 0

Answer:

Technically the big one will go down last because it's bigger but the small one will go down faster cause it's small but if not you can try it out

Alenkasestr [34]3 years ago
4 0

Answer:

they are both stones that is why

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Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,
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Energy = (power) x (time)

-- <u>For the toaster:</u>

Power = 1.4 kW  =  1,400 watts

Time = 5.4 minutes = 324 seconds

Energy = (1,400 W) x (324 s)  =  453,600 Joules

-- <u>For the CFL bulb:</u>

Power = 11 watts

Time = 10.5 hours = 37,800 seconds

Energy = (11 W) x (37,800 s)  =  415,800 Joules

-- The toaster uses energy at 127 times the rate of the CFL bulb.

-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.

-- The toaster is used for 0.0086 times as long as the CFL bulb.

-- The CFL bulb is used for 116.7 times as long as the toaster.    

-- The toaster uses 9.1% more energy than the CFL bulb.

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7 0
3 years ago
Convert the following in to SI units without changing its values<br> a. 4000 g <br>b. 2.4 km​
lana66690 [7]

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3 0
3 years ago
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answer of both parts are attached below



4 0
3 years ago
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Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

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