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Kisachek [45]
3 years ago
12

Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of

Physics
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

Explanation:

Momentum conservation

m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\

Kinetic energy conservation

\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

Solve the system

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kipiarov [429]

Answer:

This can be part of your paragraph.

Explanation:

From the cornea, the light passes through the pupil. The iris, or the colored part of your eye, controls the amount of light passing through. From there, it then hits the lens. This is the clear structure inside the eye that focuses light rays onto the retina.

4 0
2 years ago
A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force
lana66690 [7]
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2 years ago
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The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t
Andrej [43]

Answer:

ratio = 1 : 4.5

Explanation:

If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:

F_1=\frac{Gm_1m_2}{r^2}

The force F₂:

F_2=\frac{Gm_1(2m_2)}{(3r)^2}

The ratio:

\frac{F_2}{F_1}=\frac{2}{9}

8 0
3 years ago
On a cross-country trip, a couple drives 500 mi in 10 h on the first day, 380 mi in 8.0 h on the second day, and 600 mi in 15 h
Zepler [3.9K]

We know that the average speed is simply the ratio of the total distance travelled over the total duration of the trip.

total distance = 500 mi + 380 mi + 600 mi

total distance = 1,480 mi

 

total time = 10 h + 8 h + 15 h

total time = 33 h

 

So the average speed is therefore:

average speed = 1,480 mi / 33 h

<span>average speed = 44.85 mi / h</span>

8 0
3 years ago
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m
kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}

Put the value into the formula

f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}

f_{e}=3.68\ cm

Hence, The focal length of eyepiece is 3.68 cm.

3 0
3 years ago
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