Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

Answers

Answer 1

Answer:

a). The magnitude along with the direction of the electric field releasing westward force of $4.80$ × $10^(-17)$ N would be:

$3$ × $10^(-36)$ N/C is Eastward Direction

b). The magnitude along with the force of the direction that this field releases on proton would be:

$4.8$ × $10^(17) N$ in Eastward Direction

Electric Field

a). Given that,

Force $=$ $4.80$ × $10^(-17)$ N

As we know,

Force $=$ Charge × Electric Field

So,

∵ Electric Field $= Force/Charge$

$= 4.8$ × $10^(17)$ ) $/(1.6$ × $10^(-19)$ $)$

$= 3$ × $10^(36)N/C$

The direction of the field would be opposite i.e. Eastward direction due to the field carrying a -ve charge.

b). The magnitude carried by the force working on the proton would be the same with an opposite direction due to +ve charge.

∵ Force $=$ $4.80$ × $10^(-17)$ N in Eastward direction.

Learn more about "Magnitude" here:

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Answer 2

Answer:

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

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What was the main idea of Malthus theory of population

Answers

Answer:

The idea that population growth is potentially exponential while the growth of the food supply or other resources is linear.

Explanation:

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?

Answers

Answer:

Energy, $E=4.002* 10^(-19)\ J$

Explanation:

It is given that,

Wavelength of the photon, $\lambda=500\ nm=5* 10^(-7)\ m$

We need to find the photon representing the particle interpretation of this light. it is given by :

$E=(hc)/(\lambda)$

$E=(6.67* 10^(-34)* 3* 10^8)/(5* 10^(-7))$

$E=4.002* 10^(-19)\ J$

So, the energy of the photon is $4.002* 10^(-19)\ J$ . Hence, this is the required solution.

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the air outside the tank. Assume that the density of water is 1.00 g/cm3

Answers

Answer:

The net downward force on the tank is $1.85*10^(5)\ N$

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

$F=(P+\rho* g* h-P_(out))A$