As the distance between 2 objects increased, the gravitational force between them

Does the size of a paper airplane affect how far it flies

astra-53 [7]

No the only thing that affects it is how it is built

4 0

3 years ago

Part 1: Use complete sentences to explain why solar winds occur. Part 2: Give two examples in which solar winds impact Earth.

Troyanec [42]

Part 1

When the solar atmosphere accumulates a lot of magnetic energy to a point that cannot accumulate more, all that magnetic energy is suddenly released, and with it, a lot of radiation. So much, that in fact it covers all of the electromagnetic spectrum; from radio waves to gamma rays. That burst of radiation is called a solar flare. In a single solar flare the amount of radiation released is millions of times greater than all the nuclear bombs in the face if the earth exploding together. Lucky for us, most of the high-energy radiation dissipates before reaching the Earth, and the radiation that do reach us, is deflected by the Earth’s magnetic field.

Part 2

1. Not all the radiation of solar flares that reach the Earth is deflected by its magnetic field; some of them reach us and charges the upper atmosphere with ionized particles. Those particles react with the gases in the atmosphere and produce a light; that light is what we call Auroras borealis or southern nights; One the most beautiful natural spectacles in earth, who thought Auroras begin their lives as deadly solar flares.

2. Solar flares contain a lot of high-energy radiation that is extremely dangerous for our electronic devices; when they reach the Earth, they can damage sensible electronics like satellites. A very powerful solar flare could even damage all the electronic devices on the surface of the Earth.

4 0

3 years ago

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 4.5 cm. The

ANTONII [103]

Answer:

Explanation:

a) ωp = 2π radians / 1.7 s = 3.7 rad/s

b) ωs = 3.7 rad/s(9.5 cm / 4.5 cm) = 7.8 rad/s

v = (ωs)R = 7.8(65) = 507 cm/s or 5.1 m/s

c) ωs = 3.5 m/s / 0.65 m = 5.38 rad/s

ωp = 5.38(4.5 cm / 9.5 cm) = 2.55 rad/s

t = θ/ω = 2π / 2.55 = 2.463... 2.5 s

4 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers

A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

goldenfox [79]

Answer:

No
5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

211.1 = 80.5v²/11.5

30.16 = v² . . . . multiply by 11.5/80.5

5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

Comment on crossing the moat

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0

3 years ago