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bija089 [108]
3 years ago
15

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1. 2 m high. Student 2 pushes an identical box up a 5 m ra

mp from the floor to the top of the same table. Which student did the MOST work?.
Physics
1 answer:
valina [46]3 years ago
5 0

The student 2 did the MOST work because he pushed the box to a greater height.

The given parameters:

  • <em>Force applied by Student 1 = 500 N</em>
  • <em>Height of table for student 1 = 1.2 m</em>
  • <em>Height of ramp for student 2 = 5 m</em>

The work done by student 1 is calculated as follows;

Work = P.E = mgh

P.E = 1.2mg

The work done by the student 2 is calculated as;

P.E = 5mg

where;

  • <em>m </em><em>is the </em><em>mass </em><em>of the box</em>

Thus, we can conclude that the student 2 did the MOST work.

Learn more about work done here: brainly.com/question/8119756

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A lawyer drives from her​ home, located 1 mile east and 8 miles north of the town​ courthouse, to her​ office, located 4 miles w
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Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

so here we have coordinate of the office with respect to court house is given as

r_1 = (-4\hat i - 4\hat j)

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

r_2 = (1\hat i + 8 \hat j)

now the change in the position is given as the distance between office and home

d = r_2 - r_1

d = 5 \hat i + 12 \hat j

d = \sqrt{5^2 + 12^2} = 13 miles

4 0
4 years ago
Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show
ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = \frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}

We know that the inductance of the coil is given by the formula:

L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH

4 0
3 years ago
How much energy is required to raise the temperature of 5g of air by 10°C?
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Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em
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Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

         E = h f

         c= λ f

         E = h c / λ

          λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

          E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

          E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

                #_photon = P / E₀

               #_photon = 100 / 19.89 10⁻²⁰

              #_photon = 5 10²⁰ photons / s

6 0
3 years ago
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