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mart [117]
2 years ago
11

4.0 J of work are performed in stretching a spring with a spring constant of 2500 N/m. How much is the spring stretched?

Physics
1 answer:
earnstyle [38]2 years ago
3 0

energy = elastic \: potential

4 =  \frac{1}{2} kx {}^{2}

4 =  \frac{1}{2} (2500)x {}^{2}

4 = 1250x {}^{2}

x {}^{2}  =  \frac{4}{1250}  = 0.0032 \: meters

x =  \sqrt{0.0032}  = 0.05657 \: meters

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A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

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Answer:

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Explanation:

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Energy that cannot be used to do useful work is referred to as Select one: a. potential energy. b. entropy. c. kinetic energy. d
topjm [15]

In thermodynamics, entropy (symbolized as S) is a physical magnitude for a thermodynamic system in equilibrium. It measures the number of microstates compatible with the equilibrium macrostate, it can also be said that it is the reason for an increase between internal energy versus an increase in system temperature.

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Therefore the correct answer is B.

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A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the
koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as  3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of proton = 1.67 x 10⁻²⁷ kg
  • v is speed of the proton = ?
<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

Learn more about momentum here: brainly.com/question/7538238

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