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nikitadnepr [17]
3 years ago
6

At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g

ravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Engineering
1 answer:
Crank3 years ago
4 0

Answer:

1) g=31.87ft/s^2

2)m=120

W=119.64lbf

Explanation:

first part

the weight of a body with mass is calculated by the following equation

W=mg

we convert 120lb to slug

m=120lbx1slug/32.147lb=3.733slug

solving for g

g=W/m

g=119/3.733=31.87ft/s^2

second part

the mass is the same

m=120lb=3.733slug

Weight

W=3.733slug*32.05ft/s^2=119.64lbf

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What’s the number of gold atoms in a nanogram? a picogram?
zvonat [6]

Answer :

The number of gold atoms in nanogram is, 3.057\times 10^{12}

The number of gold atoms in picogram is, 3.057\times 10^{9}

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains 6.022\times 10^{23} number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-9} nanograms of gold contains \frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12} number of gold atoms

The number of gold atoms in nanogram is, 3.057\times 10^{12}

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains 6.022\times 10^{23} number of gold atoms

And, 1 grams of gold contains \frac{1g}{196.97g}\times (6.022\times 10^{23}) number of gold atoms

So, 10^{-12} picograms of gold contains \frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9} number of gold atoms

The number of gold atoms in picogram is, 3.057\times 10^{9}

8 0
3 years ago
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its
kow [346]
<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil (N_{p}) to the number of turns in its secondary coil (N_{s}) is equal to the ratio of the input voltage (V_{p}) to the output voltage (V_{s}) of the transformer. i.e

\frac{N_p}{N_s} = \frac{V_p}{V_s}            ----------------(i)

<em>From the question;</em>

N_{p} = number of turns in the primary coil = 8 turns

N_{s} = number of turns in the secondary coil = 515 turns

V_{p} = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

\frac{8}{515} = \frac{117}{V_s}

<em>Solve for </em>V_{s}<em>;</em>

V_{s} = 117 x 515 / 8

V_{s} = 7532V

Therefore, the output voltage (in V) of the transformer is 7532

6 0
3 years ago
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