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BaLLatris [955]
4 years ago
12

A current of 1.40 a flows in a wire. how many electrons are flowing past any point in the wire per second?

Physics
1 answer:
ikadub [295]4 years ago
3 0
The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of \Delta t:
I= \frac{Q}{\Delta t}

So, by using this relationship and I=1.40 A, we can find the charge passing any point in the wire in 1 second:
Q=I \Delta t=(1.40 A)(1.0 s)=1.40 C

To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron (q=1.6 \cdot 10^{-19}C):
N= \frac{Q}{q}= \frac{1.40 C}{1.6 \cdot 10^{-19} C}=8.75 \cdot 10^{18} electrons
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An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Alex Ar [27]

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

3 0
4 years ago
Read 2 more answers
What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

So

P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa

So The Force is given as:

F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N

The Force act on each side is 2.43×10⁴N

3 0
3 years ago
A girl runs and cover a distance of 200m in 20 secends.The speed of the girl is ................ m/s.
Mademuasel [1]

Speed = distance/time

Speed = (200m)/(20s)

Speed = 10 m/s

4 0
3 years ago
Find the amount of force required to move an object of 1200 kg at a velocity of 54 km/hr?​​
Mkey [24]

Answer:

0 Newtons

Explanation:

The velocity of the object does not change, it is a constant 54 km/hr. When velocity does not change, acceleration is zero. Using the formula Force = mass x acceleration, we find:

mass = 1200 kg

acceleration = 0

F  = (1200)(0) = 0

4 0
3 years ago
A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar
loris [4]

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0

5 0
3 years ago
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