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earnstyle [38]
3 years ago
6

I really need help with this, i dont know what to do first, PLEASE I NEED HELP DUE TOMORROW!!!

Physics
1 answer:
enot [183]3 years ago
5 0

Can I tell you what makes this problem so hard ?

It's having all the data WITHOUT HAVING THE STORY !

We first have to figure out what all those things are.  I mean, we don't even know what  F  is, what  d  is, what  Kef  or  Vi  is, or how  W  figures in to the whole thing.  You really have no mercy !

If my hunch is correct, the story goes like this:

-- There's an object sailing along, minding its own business, not bothering anybody, and its speed is 7.2 meters per second.

-- Somebody jumps out in front of the object and begins to push back on it with 215 Newtons of force, trying to slow it down and stop it.

-- The object is only able to go another 13 meters, pushing the guy backwards but slowing down, and then it stops.

-- The question is:  What is the mass of the object ?

Now I'll go ahead and solve the problem that I just invented:

-- Kinetic energy = (1/2) (mass) (speed²)

Before anybody touched it, the object's kinetic energy was

KE = (1/2) (mass) (7.2 m/s)²

KE = (25.92) x (mass)

-- Since that's the energy the object had, THAT's how much work the guy has to do in order to make the object stop.

Work = (force) x (distance)

Work = (215 N) x (13 meters)

Work = 2,795 N-m

-- And there you go.  The work the guy did to stop the object is the amount of energy the object had before he came along.

(25.92) x (mass of the object) = 2,795 N-m

Divide each side by 25.92:

Mass of the object = (2,795 N-m) / (25.92)

<em>Mass = 107.83 kilograms</em>

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Which wavelength of the electromagnetic spectrum are shorter than visible light and carry more energy
geniusboy [140]

Answer:

C

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A micro

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Therefore, ultravolet is shorter than visible light and carry more energy

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Explain why a real image must be produced in a camera and how the object and the lens are positioned to produce a real image whi
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/bitesize/intermediate2/physics/waves_and_optics/image_formation_from_lens/revision/1/
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4 years ago
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sp
Leokris [45]

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ q_{i} / r_{i}

where q_{i} and r_{i} are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

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What is the conversion factor between km/h and mi/h
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<span> 1 mi = 1.609 km
so X mi/hr = 1.609 * X km/hr hope this helps!!</span>
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