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Aloiza [94]
3 years ago
8

Strong winds causing a vehicle to lose all control is called

Physics
1 answer:
mixer [17]3 years ago
4 0
The answer to this question is "Buffeting". This is an unusual strong wind condition that resulted in the loss of vehicle control. This condition occurs in roads and bridges, across and along mountains that affected vehicles control. The drivers must be alert and put their full attention to overcome this condition.
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Find the pressure if a force of 2N is applied to an area of 0.00004m^2
BigorU [14]

Answer:

5 x 10^4 N/m^2

Explanation:

Pressure, force and area are related witg the following equation;

Pressure = Force /Area

From the question, we obtained the following information;

Force = 2N

Area = 0.00004m^2

Pressure =?

Pressure = Force /Area

Pressure = 2/0.00004

Pressure = 5 x 10^4N/m^2

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4 years ago
When there is no air resistance, objects of different masses (1 points)
MA_775_DIABLO [31]

Answer:

Fall at equal acceleration with similar displacements.

Explanation:

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2 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

and

m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

Then, the law of conservation of momentum demands that

m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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4 years ago
How does using a ramp to load boxes into a truck make work easier?
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"Using the ramp decreases the amount of force needed to move the boxes, but the boxes must be moved over a longer distance" is the way among the following choices given in the question that a ramp <span>to load boxes into a truck make work easier. The correct option among all the options that are given in the question is option "D".</span>
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Answer:you can go on yt tutor works

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