Is it possible to fly from Norway to USA across the North Pole?

Answers

Answer 1

Answer: Sure. The Earth is a sphere (ball-shaped), so you can follow practically
any route you want from one place to another. You could even fly from
New York to Chicago by way of the north pole.

If you want to draw a smooth curve from Oslo through the north pole to
a point in the USA, that's perfectly possible. But it's not the shortest way
to travel between those two points.

Related Questions

A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?(b Can this plane land on a runaway that is only 0.800 km long?shown work pls will reward alot of points
A 4 watt night light is left on for 8 hours each night how much work does it do per night
Determine the kinetic energy of a 1000kg roller coaster car that is moving with a speed of 20.0m/s
Direct current is the flow of an electric charge?
A tennis ball is dropped from 1.16 m above theground. It rebounds to a height of 0.866 m.With what velocity does it hit the ground?The acceleration of gravity is 9.8 m/s

In which direction should you look to see the setting moon.

Answers

Every sky object sets in the west.

It depends on the time of day. If it's just after moonrise, look in the east, low in the sky. Then near moonset, look in the west, also low in the sky.

IWhat is your mass in kilograms of you weigh 120 pounds?

Answers

One pound is (around, actually it's a little more, but you have to round it down to make the number practically usable) 0.45 kilograms, so 120 pounts is 120* 0.45, which is around 54kg.

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit of Mercury), at which point its speed is 9.5 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.)

Answers

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7* 10^(10) m$

$v_i=9.5* 10^4 m/s$

$d_2=6* 10^(12) m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$(1)/(2)mv^2_i-(GmM)/(d_1)=(1)/(2)mv^2_f-(GmM)/(d_2)$

$(1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2)=(1)/(2)v^2_f$

$v^2_f=2((1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2))$

$v_f=\sqrt{2((1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2))}$

Using the formula

$v_f=\sqrt{v^2_i+2GM((1)/(d_2)-(1)/(d_1))}$

$v_f=\sqrt{(9.5* 10^4)^2+2* 6.7* 10^(-11)* 1.98* 10^(30)((1)/(6* 10^(12))-(1)/(4.7* 10^(10)))$

Where mass of sun= $M=1.98* 10^(30) kg$

$G=6.7* 10^(-11)$

$v_f=58515.9 m/s$

What temperature is required to transfer waste heat to the environment for a heat engine to be 100 percent efficient?

Answers

This can be seen as a trick question because heat engines can typically never be 100 percent efficient. This is due to the presence of inefficiencies such as friction and heat loss to the environment. Even the best heat engines can only go up to around 50% efficiency.

Answer:

An infinite temperature

Explanation:

The efficiency of an engine is defined as:

$\eta=1 -(T_C)/(T_H)$

where

$T_H$ is the temperature at which heat enters the engine

$T_C$ is the temperature of the environment, to which the engine exhausts heat

From the formula, we see that for an engine to be 100% efficient, the fraction

$(T_C)/(T_H)$

must be equal to zero. Since the value of $T_C$ is never zero (the temperature is expressed in Kelvin, and the temperature of the environment can never be exactly 0 K), the only possibility for that to occur is that the temperature at which heat enters the engine ( $T_H$ ) is infinite, so that this fraction becomes zero and the efficiency becomes 1.

Britain and France signed an entente and became thea.
Central Forces.
c.
Central Powers.
b.
Allied Forces.
d.
Axis Powers.

Answers

im pretty sure that it is allied forces

a constant force of 391 n acts on a spacecraft of mass 7870 kg that has an initial velocity of 38 m/s. how far has the spacecraft traveled when it reaches a velocity of 4820 m/s?

Answers

Answer:

The distance travelled is 2.34*10^8 m.

Explanation:

The acceleration of the spacecraft is,

a=F/m

=391/7870 = 0.04968 m/s^2

The distance travelled,

d=v^2-v0^2/2a

=(4820)^2-(38)^2 / 2(0.04968)

=2.34*10^8 m.

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