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KengaRu [80]
3 years ago
14

Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward

each other down a long straight east-west aligned road. As the bikes travel, a bumble bee flying at a uniform speed of 25 miles per hour travels from the front wheel of one bike to the other, instantaneously reversing its course each time it encounters one of the bicycle wheels. Alas, eventually the two bicycles collide, crushing the poor bumble bee between their front wheels. What distance does the bee travel before his demise?
Physics
1 answer:
Radda [10]3 years ago
6 0

Answer:

D = 25 miles

Explanation:

To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.

Since both bicyclists collide, we know that Xa=Xb, so:

Xa = V*t = 10*t     and    Xb = 20 - V*t = 20 - 10*t

10*t = 20 - 10*t      Solving for t:

t = 1 hour  Now we can calculate the distance for the bee:

D = Vbee * t = 25 * 1 = 25 miles

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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.

_____

<em>Comment on crossing the moat</em>

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
2 years ago
On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. Which
IrinaK [193]
The correct answer for this question is this one: "The drops dripped from a bloody knife about 2 ft above the ground."
<span>On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. The statement that best accounts for the drops is that <em>the </em></span><span><em>drops dripped from a bloody knife about 2 ft above the ground.</em>
</span>
Hope this helps answer your question and have a nice day ahead.
7 0
3 years ago
The first law of thermodynamics can be given as ________.
MrRa [10]
The first law of thermodynamics can be written as
\Delta U = Q-W
where
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Q is the amount of heat absorbed by the system
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Using this form, the sign convention for Q and W becomes:
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Q < 0 --> heat released by the system (because it decreases the internal energy)
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W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)
7 0
3 years ago
Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth
barxatty [35]
An ice cube would transfer heat to another object whose temperature
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If you could find a block of solid nitrogen, its temperature would be
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6 0
3 years ago
If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of
kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass <em>m</em>₁ and <em>m</em>₂ a distance <em>r</em> apart is

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

where <em>G</em> = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

<em>F</em> = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

<em>F</em> ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)<em>g</em>, where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

<em>F</em> = <em>G m</em>₁ <em>m</em>₂ / (2<em>r</em>)² = 1/4 <em>G m</em>₁ <em>m</em>₂ / <em>r</em>²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0
3 years ago
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