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The Olympias is a reconstruction of a trireme, a type of Greek galley ship used over 2000 years ago. The power P (in kilowatts) needed to propel the Olympias at a desired speed s (in knots) can be modeled by this equation: P = 0.0289s3 A volunteer crew of the Olympias was able to generate a maximum power of about 10.5 kilowatts. What was their greatest speed? Start a New Thread

Answers

Answer 1

Answer:

Answer:

7.13559 knots

Explanation:

Maximum power = 10.5 kilowatts

$P=0.0289s^3$

where,

P = Power in kilowatts

s = Desired speed in knots

Here, P = 10.5 kW

$10.5=0.0289s^3\n\Rightarrow s^3=(10.5)/(0.0289)\n\Rightarrow s=\left((10.5)/(0.0289)\right)^{(1)/(3)}\n\Rightarrow s=7.13559\ knots$

The greatest speed of the Olympians was 7.13559 knots

10.4⁰ and 169.6⁰

Explanation:

The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;

q is the charge on the electron

v is the velocity of the electron

B is the magnetic field strength

θ is the angle that the velocity of the electron make with the magnetic field.

Given parameters

F = 1.40*10⁻¹⁶ N

q = 1.6*10⁻¹⁹C

v = 3.94*10³m/s

B = 1.23T

Required

Angle that the velocity of the electron make with the magnetic field

Substituting the given parameters into the formula:

1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ

1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ

sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶

sinθ = 1.40/7.75392

sinθ = 0.1806

θ = sin⁻¹0.1806

θ₁ = 10.4⁰

Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁

θ₂ = 180-10.4

θ₂ = 169.6⁰

Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰

Complete the calculations for total magnification produced by various combinations of the eyepiece and objective lenses. You may assume that the magnification for the eyepiece is 10X for each question. 1. When the scanning (4X) objective is used the total magnification will be:________
2. When the low power (10X) objective is used the total magnification will be:________
3. When the high power (40X) objective is used the total magnification will be:________
4. When the oil immersion (100X) objective is used the total magnification will be:_________

Answers

Answer:

a) m_ttoal = 40x, b) m_total = 100X, c) m_total = 400X,

d) m_total = 1000 X

Explanation:

La magnificación o aumentos es el incremento de del tamaño de la imagen con respecto al tamaño original del objeto, en la mayoria del os sistema optico la magnificacion total es el producoto de la magnificación del objetivo por la magnificación del ocular

m_total = m_ objetivo * m=ocular

apliquemos esto a nuestro caso

1) m_total = 4 x * 10 x

m_ttoal = 40x

2) m_total = 10X * 10X

m_total = 100X

3)mtotal = 40X * 10X

m_total = 400X

4) m _totla = 100x * 10 X

m_total = 1000 X

en este ultimo caso para magnificación grandes es decalcificar el objeto

The total magnification produced by different combinations of eyepiece and objective lenses in a microscope.

1. When the scanning (4X) objective is used, the total magnification will be 40X because the eyepiece magnification is 10X and the objective magnification is 4X.

2. When the low power (10X) objective is used, the total magnification will be 100X because the eyepiece magnification is 10X and the objective magnification is 10X.

3. When the high power (40X) objective is used, the total magnification will be 400X because the eyepiece magnification is 10X and the objective magnification is 40X.

4. When the oil immersion (100X) objective is used, the total magnification will be 1000X because the eyepiece magnification is 10X and the objective magnification is 100X.

Learn more about Microscope magnifications here:

brainly.com/question/36268796

#SPJ3

A concrete highway is built of slabs 18.0 m long (at 25 °C). How wide should the expansion cracks be (at 25 °C) between the slabs to prevent buckling if the annual extreme temperatures are −32 °C and 52 °C?(the coefficient of linear expansion of concrete is 1.20 × 10 − 5 °C-1) g

Answers

To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Original Length of the bar

$\Delta T$ = Change in temperature

$\alpha$ = Coefficient of thermal expansion

On the other hand our values are given as,

$L_0 = 18m$

$\alpha = 12*10^(-6)/\°C$

$T_2 = 52\°C$

$T_1= 25\°C$

Replacing we have,

$\Delta L = L_0 \alpha (T_2-T_1)$

$\Delta L = (18)(12*10^(-6))(52-25)$

$\Delta L = 5.832*10^(-3)m$

The width of the expansion of the cracks between the slabs is 0.5832cm

The width of the expansion cracks between the slabs to prevent buckling should be 0.5832cm.

How to calculate width?

According to this question, the following information are given:

Lo = Original length of the bar
∆T = Change in temperature
α = Coefficient of thermal energy

The values are given as follows:

Lo = 18m
T1 = 25°C, T2 = 52°C
α = 12 × 10-⁶/°C

∆L = Loα (T2 - T1)

∆L = 18 × 12 × 10-⁶ (27)

∆L = 3.24 × 10-⁴ × 18

∆L = 5.832 × 10-³m

Therefore, the width of the expansion of the cracks between the slabs is 0.5832cm.

Learn more about width at: brainly.com/question/26168065

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

In the study of sound, one version of the law of tensions is:f1= f2 √ (F1/F2)

If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.

Answers

Answer:

F1 = 80

Explanation:

f1= f2 √ (F1/F2)

Where f1 = 300, f2 = 260 and F2 = 60

Putting in the above formula

300 = 260√(F1/60)

Dividing both sides by 260

=> 1.15 = √(F1/60)

Squaring both sides

=> 1.33 = F1/60

Multiplying both sides by 60

=> F1 = 80

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?