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labwork [276]
3 years ago
10

What is the component in a car engine that would detect the need for air?​

Physics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

Control of air–fuel ratio

Oxygen sensors tell the ECU whether the engine is running rich (too much fuel or too little oxygen) or running lean (too much oxygen or too little fuel) as compared to ideal conditions (known as stoichiometric).

Explanation:

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2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
OLEGan [10]

Answer:

Relative\ Velocity = 105m/s

Explanation:

Given

V_A = 45m/s

V_B = 60m/s

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a sum\ of\ velocities\ of both trains:

This is shown below:

Relative\ Velocity = V_A + V_B

Relative\ Velocity = 45m/s + 60m/s

Relative\ Velocity = 105m/s

4 0
3 years ago
Is energy released or absorbed during the formation of a solution?
kodGreya [7K]
Energy can be released and absorbed during the formation of a solution, not one or the other. When a solute interacts with the solvent, energy is absorbed so the solvent can overcome the intermolecular bonds of the solute and energy is released, most commonly, in the form of heat, light, or a gaseous byproduct.
3 0
3 years ago
Will Give Brainliest and 25 Points
Vikentia [17]

This is the Doppler effect.

1. As the sound leaves the horn the sound waves are at first close to each other and as they move outwards they become further apart. The closer the sound waves are the louder the noise.

As the car gets the closer the sound waves get closer, so the horn becomes louder.

2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

5 0
3 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0
2 years ago
A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.
fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

5 0
3 years ago
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