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3 years ago

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The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn

ots. What forward force is exerted on the ship at this highest attainable speed? (1 knot = 0.514 m/s. MW = Megawatts)

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

$1\ knot=0.514\ m/s$

$31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s$

Power is given by

$P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N$

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

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Someone please help

saul85 [17]

Based on the attached image:

The name of the longitude line that passes through point A is the International Date Line
The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
Longitude 120° is B
Solar time at Point B is 4 pm
The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

The name of the longitude line that passes through point A is the International Date Line
The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
Longitude 120° is B
Solar time at Point B is 4 pm
the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0

2 years ago

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

5 0

3 years ago

An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which

PolarNik [594]

Answer:

R=m*g-∀fl*g*l3

Explanation:

<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>

From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced

density of iron =mass/ volume

rho=m/l3

mass=rhol3

weight fluid=rhofluid*g*Volume

weight of fluid=rhofluid*g*l3

F=∀fl*g*l3

Downward force is weight of iron

w=m*g

Reading on the spring scale

R=w-F

R=m*g-∀fl*g*l3

m=mass of iron

g=acceleration due to ravity

rhfld=density of fluid

l3=volume of fluid displaced

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3 years ago

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

or

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

What is the kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18 m/s? A.450 J B.900 J C.8,100 J D.16,000

Svet_ta [14]

KE = (1/2)·(mass)·(speed)²

KE = (1/2)·(50 kg)·(18 m/s)²

KE = (25 kg)·(324 m²/s²)

KE = 8,100 kg-m²/s²

KE = 8,100 Joules

5 0

3 years ago

Read 2 more answers