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xz_007 [3.2K]
3 years ago
6

A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.

Physics
1 answer:
Vesnalui [34]3 years ago
4 0

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

<u>Normal force:</u> It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

<u>Gravity:</u> Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

<u>Friction: </u>Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

<u>Air resistance:</u> air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

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La magnitud es de 9.8 m/s² ... la aceleración de la gravedad en o cerca de la superficie de la Tierra.

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3 years ago
A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

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Total Distance =10 km

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Total Distance =42.195 km

Total time =2.3939 h

\Rightarrow Average speed=\frac{42.195\hspace{1mm}km}{2.3939\hspace{1mm}h}=17.626\hspace{1mm}km/h

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