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2 years ago

6. The electric force between two charged objects is influenced by

Physics

1 answer:

andrew11 [14]2 years ago

3 0

Answer:

the amount of each charge and the distance between them. ... Coulomb's law states the the electric force between two charges is proportional to the product of the two charges and inversally proportional to the square distance that separates their centers.

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A metallic conductor has a resistivity of 35 × 10 −6 Ω⋅m. What is the resistance of a piece that is 20 m long and has a uniform

Nitella [24]

Answer:

Therefore the resistance of the conductor is 175Ω

Explanation:

Resistance:

Resistance of a metallic conductor is directly proportional to its length(l).

Resistance of a metallic conductor is inversely proportional to its cross section area(A).

The notation sign of resistance is R.

The unit of resistance is ohm (Ω).

Therefore,

$R \propto l$

and

$R \propto \frac{1}{A}$

$\therefore R\propto \frac{l}{A}$

$\Rightarrow R=\rho \frac{l}{A}$

ρ is the proportional constant.

It is also known as resistivity of that metal.

Given ρ=35×10⁻⁶Ω-m

l= 20 m

A= 4.0×10⁻⁶m²

$\therefore R=35\times 10^{-6}\times \frac{20}{4.0\times 10^{-6}}$

=175Ω

Therefore the resistance of the conductor is 175Ω

3 0

2 years ago

It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

TEA [102]

Answer:

$v=346.05\ m.s^{-1}$

Explanation:

Given:

initial temperature of the lead bullet, $T_i=43^{\circ}C$

latent heat of fusion of lead, $L_f=2.32\times 10^4\ J.kg^{-1}$

melting point of lead, $T_m=327.3^{\circ}C$

We have:

specific heat capacity of lead, $c=129\ J.kg^{-1}.K^{-1}$

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

$\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)$

$\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f$

$\frac{1}{2} m.v^2=m(c.\Delta T+L_f)$

$\frac{v^2}{2} =129\times(327.3-43)+23200$

$v=346.05\ m.s^{-1}$

3 0

3 years ago

calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.

Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference = 110v

Current = 20A

Unknown:

Power = ?

Solution:

To solve this problem, we use the expression below:

Power = IV

Power = 110 x 20 = 2200W

This is therefore 2.2kW

Power per hour = 2.2kWh

8 0

2 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

2 years ago

A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec

miv72 [106K]

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

7 0

2 years ago