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netineya [11]
3 years ago
11

Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

es forward at 10 m/s and car A is at rest. What fraction of the initial kinetic energy is lost in the collision?
Physics
1 answer:
Mamont248 [21]3 years ago
6 0

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

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How does the input distance of a third-class lever compare to the output distance​
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Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

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You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T
monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

     d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

     d = 1/600

     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

    sin θ = m λ / d

    θ  = sin⁻¹ (m λ / d)

First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

     θ₂ = 51.81 °

3 0
3 years ago
Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?
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Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult  to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of  material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout  the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

6 0
4 years ago
A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov
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Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ =  3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

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hence, the total momentum is 8 .2 kg m/s in north direction.

7 0
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