What is the scientific definition of tracked?

Hopping is performed on 2 feet<br><br><br> True<br> False

True welcome₩₩ sorry

8 0

2 years ago

How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00

klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is $9 \times 10^{-4} m^{2}$ . As the capacitance of capacitor is given as follows.

C = $\frac{\epsilon_{o}A}{d}$

It is known that the dielectric strength of air is as follows.

E = $3 \times 10^{6} V/m$

Expression for maximum potential difference is that the capacitor can with stand is as follows.

dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

Q = CV

= $\frac{\epsilon_{o} A}{d} \times E \times d$

= $\epsilon_{o}AE$

= $8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}$

= $1.062 \times 10^{-8} C$

or, = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0

3 years ago

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge

Troyanec [42]

Answer:

$I_{l} =44.84 \mu A$

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire = 1.30×10⁻² Ω

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

$\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))$

$\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{V}{R}$

$I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}$

$I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))$

$I_{l} =44.84 \times 10^{-6}\A$

$I_{l} =44.84 \mu A$

3 0

3 years ago

A quantity that has only one magnitude

Len [333]

Answer:

scalar quantity

Explanation:

Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude.

3 0

3 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

11 months ago