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4 years ago

Formula of relative displacement

Physics

1 answer:

Zolol [24]4 years ago

8 0

what is the question?

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Anit [1.1K]

Answer :D

Some characteristics shared by all electromagnetic waves are that they all travel at the same speed of light and their own transmission does not need medium. These wave types can also travel through empty spaces

6 0

3 years ago

Read 2 more answers

A rubber rod rubbed with fur acquires a charge of -4,8×10(-9)C. What is the charge on the fur? How much mass is transferred to r

mote1985 [20]

1) The charge left on the fur is equal and opposite to the charge transferred to the rod:
$Q=+4.8 \cdot 10^{9} C$
In fact, when the rod is rubbed with the fur, a net charge of $Q=-4.8 \cdot 10^{-9} C$ has been transferred to the rod, leaving it negatively charged. If we assume the fur was initially neutral, this means that we have now an excess of positive charges on the fur, and the amount of this charge must be equal (in magnitude, but with opposite sign) to the charge transferred to the rod.

2) The mass transferred to the rod is equal to the total mass of the electrons transferred to the rod.
The charge transferred to the rod is
$Q=-4.8 \cdot 10^{-9} C$
The charge of 1 electron is
$e=-1.6 \cdot 10^{-19} C$
So the number of electrons transferred is
$N= \frac{Q}{e}= \frac{-4.8 \cdot 10^{-9} C}{-1.6 \cdot 10^{-19} C}=3.0 \cdot 10^{10}$

The mass of 1 electron is $m=9.1 \cdot 10^{-31} kg$ , therefore the total mass transferred to the rod is
$M=Nm=(3 \cdot 10^{10})(9.1 \cdot 10^{-31} kg)=2.73 \cdot 10^{-20} kg$

7 0

3 years ago

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

A computer is purchased for $2816 and depreciates at a constant rate to $0 in 8 years. Find a formula for the value, V , of the

marusya05 [52]

Answer:

The formula its $f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$
After 5 years, the computer value its $ 1056

Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

Starts at 2816. $f(0 \ years) \ = \ \$ \ 2816$
Reach 0 at 8 years. $f( 8 \ years) \ = \ \$ \ 0$
Depreciates at a constant rate. m

We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

$f(t) \ = \ m\ t \ + \ b$ ,

where m its the slope of the line and b give the place where the line intercepts the y axis.

So, we can use this formula with the data from our problem. For the first condition:

$f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816$

$b = \$ \ 2816$

So, b = $ 2816.

Now, for the second condition:

$f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0$

$m \ (8 \ years) = \ - \$ \ 2816$

$m = \frac{\ - \$ \ 2816}{8 \ years}$

$m = \ - \ 352 \frac{\$ }{years}$

So, our formula, finally, its:

$f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$

<h3>After 5 years</h3>

Now, we just use t = 5 years in our formula

$f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816$

$f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816$

$f(5 \ years) \ = $ \ 1056$

4 0

4 years ago

Help, please

kvv77 [185]

Answer:

I am going to answer the questions, according to the numbers of empty spaces to fill,1.electric charge,2.force,3.field lines,4.negatives,5.positive,6.affected,7.attract,8.repel.

Explanation:

if u read it filling in the spaces using this answers,u will understand

5 0

2 years ago

Formula of relative displacement​

Formula of relative displacement