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3 years ago

A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s

Physics

1 answer:

Romashka [77]3 years ago

6 0

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

p₀ = m v₀

celestines que clash case

p_f = (m + M) v

po = pf

m v₀ = (n + M) v

v = $\frac{m}{m+M}$

calculemos

v= $\frac{0.200}{0.200+M} 0.450$

v= 0.09 m/s

elastic shock case

p₀ = m v₀

p_f = m v₁ +M v₂

p₀ = p_f

m v₀ = m v₁ + m v₂

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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

The denizens of a distant planet live under an atmosphere made predominantly of CO2 which is a dispersive medium while attending

LuckyWell [14K]

Since the frequency of sound in a medium is constant, therefore, the concert-goers would hear the low notes and high notes at the same time.

<h3>What is a dispersive medium?</h3>

A dispersive medium is a medium which spreads out or disperses a substance passing through it.

Since CO2 is a dispersive medium, it means sound waves passing through it would be dispersed based on wavelength.

The note of a sound depends on its frequency, the higher the frequency, the higher the note.

Frequency of sound is constant, therefore, the concert-goers would hear the low notes and high notes at the same time.

Learn more about dispersion of sound at: brainly.com/question/781734

5 0

2 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago

Explain how heat (thermal energy) can change the motion (kinetic energy) of the particles in objects.

Vika [28.1K]

Answer:

Adding heat makes the particles move faster so the particles have more kinetic energy when more thermal energy is added

Explanation:

4 0

3 years ago

Identify the pros and cons of using mirrors versus lenses in each of the following applications:

disa [49]

Answer:

Explanation:

A lens – An object, usually made of glass, that focuses or defocuses the light that passes through it.

A mirror – A smooth surface, usually made of glass with reflective material painted on the underside, that reflects light so as to give an image of what is in front of it.

Telescope – monocular optical instrument possessing magnification for observing distant objects, especially in astronomy.

Refracting telescopes – are telescopes that use lenses are and those that use concave parabolic mirrors are called reflecting telescopes.

Pros of mirror telescope

• They are easier to construct and not expensive to produce

• made larger and more durable (more light can be directed to the eyepiece which is good)

• cannot have any occlusions

• mirrors have less spherical aberration

• reflect all wavelengths of light equally

• more magnification power for cheaper version

Cons of mirror prisms

• it must also be realigned after cleaning, which can be expensive - maintenance disadvantage

• Reflective Surface Disadvantage

Pros of Lenses telescope

• Easy to use

• More reliable

Cons of Lens telescope

• are not easy to construct and expensive to produce

• May have occlusions

• spherical aberration are more

• Doesnt reflect wavelengths of light equally they bend light differently

• Less magnification power for cheaper version

• heavy

• Longer body

Pro of Mirror periscope

• it helps us to see further than our view that is what is above us or sometimes even below us.

•Light weight

disadvantage:

• it may not work properly and show distorted images due to fog. ; Some have a narrow field of view ; May be detected by others Ungainly (long, takes up space),

•Not as rugged as prisms.

• Maintenance of reflection surface

Pros of Prism periscopes

• incorporated lenses for magnification and function as telescopes.

• They typically employ prisms and total internal reflection instead of mirrors, because prisms, which do not require coatings on the reflecting surface, are much more rugged than mirrors.

• May be fitted with additional optical capabilities such as range-finding and targeting.

Cons.

•mechanically disadvantage: Complex optically ; Some have a narrow field of view ; May be detected by others

Ungainly (long, takes up space)

• Heavy weight

Lens - Terrestrial telescope

Pro: An advantage is that it makes it possible to vary the magnification of the telescope.

Cons- This system has the disadvantage increasing the length of the telescope.

Mirror- Terrestrial telescope (Cassegrain)

Pro: Good for distant terrestrial viewing.

greater magnification is attained.

They are more shorter

Cons- It is not what people expect a telescope to look like.

Slight light loss due to secondary mirror obstruction compared to refractors. Generally not suited for most terrestrial applications nor for to view objects in the sky.

6 0

3 years ago