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3 years ago

Earth orbits the sun once every 365.25 days. Find the average angular speed of earth about the sun. Answer in units of rad/s.

Physics

1 answer:

Talja [164]3 years ago

7 0

Answer:

2.0x10-7 rad/s.

Explanation:

Calculating the angular velocity of the Earth is a deceptively easy task. The reason for this is simple - the angular velocity is defined as the angle subtended in a certain time.

We know the Earth goes round the Sun, all the way around is 2 radians (360 degrees). We also know that it takes a year (approx 365 days) which is therefore about 3.2x107 secs.

Therefore = 2 / 3.2x107 = 2.0x10-7 rad/s. We have calculated the angular velocity.

However, if we can measure the distance to the Sun we can also calculate the velocity of the Earth relative to the Sun. Although unless we define a direction this is more technically known as the speed. This can be done by looking at the definition of the radian. The radian is a unit which conects the radius of an arc, the length of the arc and the angle subtended by the arc. The formula for this is s = r x (where s is the length of the arc, r is the radius and the angle). So if we know the radius of the Earth's orbit (1.5x10¹¹m) we can substitute the angular velocity from our previous equation to give v = x r (where v is the velocity, the angular velocity and r the radius).

So, the Earth travels through space (relative to the Sun) at: v = 2.0x10-7 x 6.4x106 = 3.0x104m/s

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Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li

Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

$L=L_0\sqrt{1-\frac{v^2}{c^2} }$

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

$L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }$

= 1.136×10^17 m

L= $= \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}$ m

so L= 12 light years

8 0

4 years ago

A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg , acting in upward direction

T - mg = m a

T = mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0

3 years ago

PLS SHOW WORK !!!!!!! ASAP

Lynna [10]

Answer:

1- For the track B. The potential energy is the same for the two cars, but because of the slope of the track, the car B earn kinetic energy faster. The gravitation acceleration of the cars will be g•sinθ, and the angle of the track B will have a bigger value for sinθ

2- The conservation of energy applies because the roller coaster is a closed track. When a car climb the track, it earn GPE, which is given by mgh, when it get down in the track, it transform GPE in KE, which is given in 1/2mv².

Position of car (m) GPE KE GPE + KE

top (30m) 60000 0 60000

bottom (0m) 0 60000 60000

halfway down (15m) 30000 30000 60000

three-quarters way down 15000 45000 60000