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olga nikolaevna [1]
3 years ago
8

In a fluorescent tube of diameter 4.8 cm , 2.7 × 1018 electrons (with a charge of −e) and 2.4 × 1018 positive ions (with a charg

e of +e) flow in opposite directions through a cross-sectional each second. What is the current in the tube? The fundamental charge is 1.602 × 10−19 C. Answer in units of A.
Physics
1 answer:
Lesechka [4]3 years ago
7 0

Answer: 0.817A

Explanation:

Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It

It means a current of one A flow in the opposite direction.

This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,

In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.

According to this, the current in the given problem is

[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s

= 0.817 A

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Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and
Zigmanuir [339]

Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

Intensity I

I= 1.39\ KW/m^2

Speed\ of \ light = 2.99\times 10^8\ m/s

Radius of earth,R = 6370 Km

We know that surface area of earth, A

A=\pi R^2

A=\pi (6370\times 10^3)^2\ m^2

A=1.27\times 10^{14}\ m^2

As we know that pressure due to intensity given as

P=\dfrac{I}{V}

V =Velocity of light

V=3\times 10^8\ m/s

P=\dfrac{1.39\times 1000}{=3\times 10^8}

P=4.6\times 10^{-6}\ Pa

We know that force F

F = P .A

F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

F=5.8\times 10^{8}\ N

b)Gravitational force F

F=G\dfrac{m.M}{r^2}

M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

G =6.67\times 10^{-11}Nm^2/kg^2

So F

F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

7 0
3 years ago
Read 2 more answers
During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
If the ankylosaurs starts running and accelerates at 1.3m/s^2 for 3 seconds how far does he make it before the velociraptor catc
sveta [45]

Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.

Acceleration = 1.3 m/s²

Velocity: ∫ 1.3 dx = 1.3x + c m/s

Distance: ∫ 1.3x dx = 1.3x²/2 + c m

Distance run: 1.3*3²/2 = 5.85 m

<em>What</em><em> </em><em>bad</em><em> </em><em>thing</em><em> </em><em>happened</em><em>?</em>

7 0
2 years ago
Which of the following is a true statement?
pishuonlain [190]

I think the answer is A.........

7 0
2 years ago
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
2 years ago
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