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Salsk061 [2.6K]
3 years ago
8

consider the following controls in an automobile in motion : gas pedal , brake , steering wheel . what are the controls in this

list that cause an acceleration of the car ?
Physics
2 answers:
kenny6666 [7]3 years ago
5 0

Answer: Gas pedal

Explanation: Gas pedal is also known as accelerator pedal.

It is a trigger in an automobile which accelerates the process of fuel combustion in the engine. Automobiles consist of an Internal Combustion engine i.e. the fuel burns inside a closed chamber to generate the energy.

When the gas pedal is triggered the supply of the fuel to the engine increases and the rate of fuel combustion also increases which in turn produces greater power. And we know that power is the rate of doing work or the rate of energy transfer which causes the acceleration in the vehicle.

tatyana61 [14]3 years ago
3 0
All of the controls can cause acceleration. The gas pedal and brake cause the car to change speed and the steering wheel can cause a change in velocity by changing its direction which is also acceleration. 
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A 19 nC charge is moved in a uniform electric field. The electric field does 5.3 μJ of work as the charge moves from point A to
Marizza181 [45]

Answer:

The potential difference between points A and B is 278.95 volts.

The potential difference between points B and C is -642.10 volts.

The potential difference between points A and C is -363.15 volts.

Explanation:

Given :

Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C

Work is done to move a charge from point A to B, W₁ = 5.3 μJ

Work done to move a charge from point B to C, W₂ = -12.2 μJ

Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.

The relation between work done and potential difference is:

W = qV  

V = W/q    ....(1)

Using equation (1), the potential difference between points A and B is:

V_{1}=\frac{W_{1} }{q}

Substitute the suitable values in the above equation.

V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }

V₁ = 278.95 V

Using equation (1), the potential difference between points B and C is:

V_{2}=\frac{W_{2} }{q}

Substitute the suitable values in the above equation.

V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }

V₂ = -642.10 V

The potential difference between points A and C is:

V₃ = V₁ + V₂

V₃ = 278.95 - 642.10

V₃ = -363.15 V

8 0
3 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?
Lisa [10]
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
7 0
2 years ago
A car travels 200km in 3.0 hours. Determine the average velocity of the car
Masja [62]

Answer:

<h2>66.67 km/hr</h2>

Explanation:

The average velocity of the car can be found by using the formula

a =  \frac{d}{t }  \\

d is the distance

t is the time taken

From the question we have

a =  \frac{200}{3}  \\  = 66.66666...

We have the final answer as

<h3>66.67 km/hr</h3>

Hope this helps you

4 0
2 years ago
Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
2 years ago
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