In order for work to be done a___ must be exerted on an object causing it to__

A circuit component with a high resistance will have a low electric current.

Zepler [3.9K]

You haven't said what 'high' resistance or 'low' current means, so there's way not enough info to nail the statement as true or false. The most precise answer is "certainly could be but not necessarily". Anyway, the current in the circuit depends on BOTH the resistance AND the voltage. So without knowing the voltage too, you can't say anything about the current.

6 0

3 years ago

A ball of mass 5 kg attached to a string is swung in a horizontal circle of radius 0.5 m. If the tension in the string is 10 N,

kirill115 [55]

Answer:

0 J

Explanation:

given,

mass of the ball = 5 kg

radius of the horizontal circle = 0.5 m

tension in the string = 10 N

Work done = ?

Work done by the tension in the circular path will be equal to zero.

This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.

so, work done = F s cos θ

θ = 90°,

work done = F s cos 90° ∵ cos 90° = 0

Work done = 0 J

8 0

3 years ago

If the cross-section of a wire of fixed length is doubled, how does the resistance of that wire change? (this is for studying fo

Ymorist [56]

If the cross-section of a wire of fixed length is doubled, the resistance of that wire change into doubled.We know that the total length of the wires will affect the amount of resistance. The longer the wire, the more resistance that there will be so the answer is doubled.

6 0

3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$