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3 years ago

A current of 1.70 A flows in a wire. How many electrons are flowing past any point in the wire per second

Physics

1 answer:

castortr0y [4]3 years ago

8 0

Answer:

1.0625×10¹⁹ electrons

Explanation:

From the question,

Using,

Q = It........................ Equation 1

Where Q = charge, I = current flowing in the wire, t = time (seconds)

Given: I = 1.70 A, t = 1 Seconds.

Substitute these values into equation 1

Q = 1.7(1)

Q = 1.7 C.

But,

1.6×10⁻¹⁹ C = 1 electron

Therefore,

1.7 C = 1.7/(1.6×10⁻¹⁹) electrons

= 1.0625×10¹⁹ electrons

Hence, 1.0625×10¹⁹ electrons flows through the wire per seconds.

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What makes a magnet on the atomic level?

yanalaym [24]

Answer:

Magnetism at atomic /sub - atomic levels is mostly due to charged particles called electrons. Electrons have spin which give them angular momentum and thus a magnetic moment associated with it. That is the cause of Magnetism at atomic levels. Electrons fill up orbitals in atoms in pairs.

4 0

3 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

$\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s$

4 0

3 years ago

What is the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m?

notsponge [240]

Answer:

a ) 24 m/s

Explanation:

Given,

Frequency ( f ) = 6 Hz

Wavelength ( λ ) = 4 m

To find : Speed ( v ) = ?

Formula : -

v = f x λ

= 4 x 6

= 24 m/s

Therefore, the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m

is 24 m/s.

4 0

3 years ago

A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp

Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

$\frac{pV}{T}=nR$

For a fixed amount of gas, n is constant, so we can write

$\frac{pV}{T}=const.$

Therefore, for a gas which undergoes a transformation we have

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

$p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K$