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FromTheMoon [43]
3 years ago
6

Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

top in 70 feet. If the van is traveling 48 miles per​ hour, what is its stopping​ distance?
Physics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

v_f^2 - v_i^2 = 2 ad

so here we have

0^2 - 40^2 = 2 a (70 feet)

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

0^2 - 48^2 = 2 a (d)

now divide the above two equations

\frac{40^2}{48^2} = \frac{70 feet}{d}

d = 100.8 ft

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The Milky Way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to Hubble
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Answer: SBb

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So, according to this classification, the Milky Way is a barred spiral galaxy (SBb in Hubble's notation system) because it has a central bar-shaped structure of bright stars that spans from one side of the galaxy to the other. In addition, its spiral arms seem to emerge from the end of this "bar".

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A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc
vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 6,000 kg is the mass of the first train

u_1 = 10 m/s is the initial velocity of the first train

m_2 = 4,000 kg is the mass of the second train

u_2 = 0 is the initial velocity of the second train  (initially at rest)

v is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s

2)

There are two types of collisions:

  • Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
  • Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J

After:

K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

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brainly.com/question/2370982  

brainly.com/question/9484203  

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