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12345 [234]
3 years ago
8

During a collision, a bus runs into a skateboarder inelastically. If the mass of the skateboarders is 70 kg and is moving at 12

m/s to the right, what is the final velocity of the skateboarder if the 2000 kg bus is traveling at 25 m/s to the left before the collision?
Physics
1 answer:
agasfer [191]3 years ago
5 0

The final velocity of the skateboarder is 23.75 m/s.

We can use conservation of momentum to calculate the magnitude of the final velocity.

Momentum before collision is equal to the momentum after collision

m1v1+m2v2=(m1+m2)v

here, v is final velocity of the skateboarder, assuming right side velocity as positive, and left side velocity as negative.

70*12+2000*(-25)=(70+2000)v

v=-23.75 m/s

Here, negative sign indicates that both skyboarder and bus will move in left direction.

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In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
3 years ago
What is this? Plz I need help...
RSB [31]

Answer:

the time it takes for one complete back and forth swing

Explanation:

the Mark's is showing you the time it swings back and forth

4 0
3 years ago
A plane flying against a jet stream will travel faster than a plane traveling with a jet stream. Please select the best answer f
stepan [7]

"with the wind" is a tail-wind, and the speeds are added to get the groundspeed.

"against the wind" is a head-wind, and the windspeed is subtracted from the airspeed.

3 0
3 years ago
Read 2 more answers
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
3 years ago
A magnifying glass or a hand lens is a bi-convex lens. It is convex on both sides, meaning that the glass is curved outward to
tankabanditka [31]

Answer:

Refracted

Explanation:

If on UsaTestPrep

7 0
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