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3 years ago

6

How are frequency and were peroid related?

1 answer:

hodyreva [135]3 years ago

7 0

Answer:

Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity.

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The pan because it is directly on there to absorb heat while the handle is there more towards the end and will have heat reaching the start of the handle while the metal is already hot

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1. Charges acquired by rubbing is called_____

Leokris [45]

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static electricity and then lightning rod

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Shawn and his bike have a total mass of

Trava [24]

Answer:

Shawn's kinetic energy is 61.45 J.

Explanation:

Given;

Mass of the system is, $m=52.3\ kg$

Displacement of the bike is, $d=1.6\ km=1.6\times 1000=1600\ m$

Time taken is, $t=17.4\ min=17.4\times 60=1044\ s$

Let the constant velocity be $v$ .

For constant velocity, magnitude of velocity is given as distance by time.

Therefore, $v=\frac{d}{t}=\frac{1600}{1044}=1.533\ m/s$

Now, kinetic energy of a body is given as:

$KE=\frac{1}{2}mv^2$

Here, $m=52.3\ kg,\ v=1.533\ m/s$

$\therefore KE=\frac{1}{2}\times 52.3\times (1.533)^2\\KE=0.5\times 52.3\times 2.35\\KE=61.45\ J$

Therefore, Shawn's kinetic energy is 61.45 J.

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4 years ago

Two large, parallel, conducting plates are 1.00 cm apart and having electric charges of equal magnitudes (80.0 nC/m^2) and oppos

boyakko [2]

(a) 9040 V/m

The magnitude of the electric field between two parallel, oppositely charged plates is given by

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the charge surface density on each plate

$\epsilon_0 = 8.85\cdot 10^{-12}$ is the vacuum permittivity

In this problem, the magnitude of the charge density on each plate is

$\sigma = 80.0 nC/m^2 = 80.0\cdot 10^{-9} C/m^2$

Substituting into the formula, we find the magnitude of the electric field:

$E=\frac{80.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=9040 V/m$

(b) 90.4 V

The electric potential difference between the two plates is given by

$\Delta V=Ed$

where

E is the magnitude of the electric field

d is the separation between the two plates

In this problem, we have

E = 9040 V/m

d = 1.00 cm = 0.01 m

Substituting into the formula, we find

$\Delta V=(9040)(0.01)=90.4 V$

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Arte-miy333 [17]

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