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Romashka [77]
4 years ago
5

What is/are the external cause code(s) for a passenger involved in a mva that lost control on the highway and hits a guardrail?

Physics
1 answer:
Sholpan [36]4 years ago
6 0
Here is the answer. The external cause code for a passenger that is involved in a MVA that lost control on the highway and hits a guardrail is E815. Your fifth digit is a 1 for a motor vehicle passenger. You would use an additional E code when a place of occurrence (example, home or parking lot) is documented. In this case, the location is documented as the highway. Hope this helps. 
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almond37 [142]

Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

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Many of water's emergent properties, such as its cohesion, its high specific heat, and its high heat of vaporization, result fro
Bingel [31]

Answer:

Option 5.

Explanation:

Many of the properties of water like high specific heat, cohesion, high vaporization heat, etc can be contributed to the polar nature of water molecule.

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3 years ago
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salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

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So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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