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Leona [35]
3 years ago
7

A 75 kg Spider Man running at 3.0 m/s jumps onto a trash can lid that has a mass of 10kg and that is already moving in the same

direction as Spiderman, at 2.0 m/s. What will be the final velocity of Spiderman and the lid assuming that the lid is on a level slick surface and that there is no friction?
Physics
2 answers:
QveST [7]3 years ago
6 0
In order to find the final velocity of the skier and the trash can lid, we may apply the principle of conservation of momentum, which states that the total momentum of a system remains constant. Mathematically, in this case:
m₁v₁ + m₂v₂ = m₃v₃
Where m₃ and v₃ are the combined mass and velocity.

75*3 + 10*2 = (75 + 10)*v₃
v₃ = 2.88 m/s

The final velocity is 2.88 m/s
nikdorinn [45]3 years ago
4 0

The final velocity is 2.88 m/s

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man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
Define the Element of Art, value. If you were using a pencil, how would you create a darker value?
ankoles [38]

Elements of art are stylistic features that are included within an art piece to help the artist communicate. The seven most common elements include line, shape, texture, form, space, colour and value, with the additions of mark making, and materiality.

You would create a darker value if you shaded it, shading it makes it darker.

8 0
3 years ago
The strongest ernergy wave in the EM Spectrum is 'microwave'<br> true<br> false
Ratling [72]
Pretty sure it’s false....................
6 0
3 years ago
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