3 years ago

How fast is the car traveling if it has driven a total of 200 miles in 5.5 hours?

1 answer:

6 0

Average speed = (distance covered) / (time to cover the distance)

   = (200 miles) / (5.5 hours)

   = (200/5.5) mile/hour

   = 36.4 miles per hour (rounded)

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3 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

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because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

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3 years ago

A portion of the mesosphere and thermosphere known for its ability to "bounce" radio signals is the _____________

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in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

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The initial speed of the vehicle before the collision is 8 m/s.

Let the mass of the vehicle = m
Let the initial speed of the vehicle stopped = u
The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>

The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s$

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

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2 years ago

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3 years ago

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