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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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What is the magnitude of velocity for a 2000 kg car possessing 3000 kgm/s of momentum
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For this case we have that by definition, the momentum is given by:

p = mv

Where,

  • <em>m: mass </em>
  • <em>v: speed </em>

Therefore, replacing values we have:

3000 = (2000) v

From here, we clear the value of the speed:

v = \frac {3000} {2000}\\v = 1.5 \frac {m} {s}

Answer:

The magnitude of velocity is:

v = 1.5 \frac {m} {s}

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3 years ago
In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
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