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AnnyKZ [126]
2 years ago
6

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte

rval?
Physics
1 answer:
musickatia [10]2 years ago
6 0

Answer:

Number of electrons, n=5.62\times 10^{21}

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, I=\dfrac{q}{t}

I\times t=n\times e

n=\dfrac{It}{e}

e is the charge of an electron

n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}

n=5.62\times 10^{21}

So, the number of electrons pass through the resistor is 5.62\times 10^{21}. Hence, this is the required solution.

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ivanzaharov [21]

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²


6 0
2 years ago
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Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c
Rus_ich [418]

Answer:

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

5 0
2 years ago
A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr
const2013 [10]

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

7 0
2 years ago
A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0
GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

F=qE

Put the value into the formula

F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}

We need to calculate the total charge on the rod

Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}

Put the value into the formula

Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}

Q=47.8\times10^{-9}\ C

Q=47.8\ nC

Hence, The total charge on the rod is 47.8 nC.

6 0
2 years ago
For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a
jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

R=\frac{L}{A\sigma }

R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

i = \frac{V}{R}

i = \frac{0.060}{0.0178}

i = 3.4 A

(c)

Current density is given as

J = \frac{i}{A}

J = \frac{3.4}{5.3\times10^{-6}}

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

E = \frac{J}{\sigma }

E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}

E = 9.01 x 10⁻³ V/m

5 0
2 years ago
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