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lys-0071 [83]
2 years ago
8

Predict the energy transformation that occurs when carson sands a piece of wood with sandpaper.

Physics
1 answer:
tigry1 [53]2 years ago
6 0

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

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A box contains about 5.54 1021 hydrogen
Aleks [24]

Answer:E=33.71\ J

Explanation:

Given

No of atoms of hydrogen N=5.54\times 10^{21}

Temperature of room T=21^{\circ}\approx 294\ K

Thermal energy of the atoms is given by

E=\frac{3}{2}NkT

where k=boltzmann constant

E=\frac{3}{2}\times 5.54\times 10^{21}\times 1.38\times 10^{-23}\times 294

E=33.71\ J

Hence the energy of the atoms is 33.71\ J

5 0
3 years ago
(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
3 years ago
An object is placed 100 cm in front of a diverging lens of focal length -25cm. A converging lens of focal length 33 1/3 cm is pl
bagirrra123 [75]
We use 1/o + 1/i = 1/f  where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>

<span>For the case of the problem,</span>

<span>o = (20 + 30)  = 50 cm </span>

<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>

<span>M = magnification = - i / o </span>

<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>

<span>m2 = -100/50 = -2 </span>

<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
8 0
4 years ago
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10
Mars2501 [29]

Answer:

a) a=5.7551 \times g

b) d=20.5539\times g

Explanation:

Given:

  • speed of rocket initially, v_i=0\ m.s^{-1}
  • top speed of rocket after acceleration, v=282\ m.s^{-1}
  • time taken to get to the top speed, t_i=5\ m.s^{-1}
  • final speed of the rocket, v_f=0\ m.s^{-1}
  • time taken to get to the final speed after reaching the top speed, t_f=1.4\ s

Now the acceleration:

a=\frac{v-v_i}{t_i}

a=\frac{282-0}{5}

a=56.4\ m.s^{-2}

Now as a fraction of gravity:

a=\frac{56.4}{9.8}\times g

a=5.7551 \times g

Now, the deceleration:

d=\frac{0-282}{1.4}

d=201.4285\ m.s^{-2}

Now as a fraction of gravity:

d=\frac{201.4285}{9.8}\times g

d=20.5539\times g

6 0
4 years ago
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