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4 years ago

How does energy change (transforms) as the mass is dropping?

Physics

1 answer:

olganol [36]4 years ago

4 0

Answer:

Mass has total mechanical energy, which is the sum of kinetic and potential energy. as the mass is dropping, potential energy is converted into kinetic energy so mechanical energy is preserved If there is no friction. If there is friction, some of the mechanical energy is lost as heat energy so it changes.

Explanation:

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A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

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4 years ago

Two beams of coherent light are shining on the same sheet of white paper. when referring to the crests and troughs of such waves

tensa zangetsu [6.8K]

Middle of the paper

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4 years ago

The critical angle for a light ray travelling from a ruby into ethyl alcohol is 50.6 degrees celsius. Determine the index of ref

stepan [7]

In order to answer this question, I realized that I needed to know the index
of refraction for ruby, so I went and looked it up. It's 1.762 to 1.770 .

I started trying to remember how to use this number and the critical angle
to find the index of refraction of the other medium. That's when I saw the
absurd unit "degrees celsius" for the critical angle, and I got discouraged.
But I perked up very quickly, when I realized that I'm still on the "index of
refraction" list, and while I'm there, I might as well just go ahead and
look up ethyl alcohol too.

It's 1.36 .

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3 years ago

You just tested the hypothesis a proposed model for the position of bright fringes produced by two slits. Explain what the model

Viefleur [7K]

Answer:

two-slit interference model was proposed by Young d sin θ = m λ

Explanation:

The two-slit interference model was proposed by Young, it establishes that if a coherent source of light passes through two slits, the shape of the given pattern is a consequence of the relative phase difference between the two rays; mathematically it can be expressed by

d sin θ = m λ

m= 0, 1, 2, 3, ...

for constructive interference, that is, the two rays arrive with a number between wavelengths.

D is the distance between the slits, tea the angle between the two rays, m an integer and m the wavelength used.

In a simulation a pattern of slits of equal intensity and equally spaced is observed.

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3 years ago

The mass of the sphere is 0.151 kg. What is the rotational inertia of the wheel

shusha [124]

Complete Question:

The mass of the sphere is 0.151 kg. What is the rotational inertia of the wheel? Recall that the radius of the sphere is about 0.312 m. Express your answer with three significant figures in Kg.m².

Answer:

I = 5.88*10⁻³ kg.m²

Explanation:

It can be showed that the rotational inertia (or moment of inertia) for a solid sphere of radius r and mass m, regarding any axis passing through the center of the sphere, can be expressed as follows:

I = (2/5)*m*r² (1)

where m= 0.151 kg, and r =0.312 m. (Our givens).

Replacing in (1) we have:

I = (2/5)*0.151 kg*(0.312m)² = 5.88*10⁻³ kg.m²

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4 years ago

How does ​energy change (transforms) as the mass is dropping?

How does energy change (transforms) as the mass is dropping?