Tony ran 600 meters in 60 seconds. What was Tony's speed during the<br> race?

1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit

Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:

Given as

y = nDλ/w Eqn 1

where

w = width of slit

D = distance to screen

λ = wavelength of light

n = order number

Making x the subject of the formula gives,

w = nDλ/y

Given

y = 0.0149 m

D = 0.555 m

λ = 588 x 10-9 m

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1, y = nDλ/w

given, D = 27cm = 0.27m

λ = 632 x 10-9 m

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0

2 years ago

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 batt

Natali [406]

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:

a

The additional resistance is $R_z = 4.4 \Omega$

b

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

c

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

d

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

e

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

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

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

$C = 2 * 6 * 4$

$C = 48 W$

6 0

3 years ago

Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.

SVEN [57.7K]

The strength of electric field E is 17 N / C.

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<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

E = F / q

where E represents the electric field strength,

F represents the force in newton,

q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

force F = 5.0 N

Electric field strength E = force / charge

= 5.0 / 0.30

E = 16.66 = 17 N / C.

5 0

2 years ago

Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the

Whitepunk [10]

The twice as heavy weight will hit the ground with more force, or impact.

4 0

2 years ago

Read 2 more answers

A ball is thrown straight up with a speed of 30 m/s, and air resistance is negligible. How long does it take the ball to reach t

PolarNik [594]

Answer:

3 seconds

Explanation:

Applying,

v = u±gt................ Equation 1

Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.

From the question,

Given: v = 0 m/s ( at the maximum height), u = 30 m/s

Constant: g = -10 m/s

Substitute these values into equation 1

0 = 30-10t

10t = 30

t = 30/10

t = 3 seconds

6 0

2 years ago

Tony ran 600 meters in 60 seconds. What was Tony's speed during the race?