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Kryger [21]
3 years ago
6

With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c

oncept of an ultralight car. Using carbon composites, lighter steels, and plastics, a fuel-efficient car can be manufactured at 540 kg. How much less does an ultralight car weigh compared to a 1450-kg Honda Accord (2007)
Physics
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

            Where: m is the object's mass

                         g is the  gravitational acceleration in the surface earth

                         g = 9.8 m/s2

The the ultralight car's weight is:

w_{uc} = (540)(9.8)

w_{uc} = 5292 N

And the Honda Accord's weight is:

w_{HA} = (1450)(9.8)

w_{HA} = 14210 N

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

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There is little vertical air movement in the __ because of the cold, heavy air at the bottom of the layer.
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The correct answer is the Mesosphere. <span>Mesosphere </span><span>is the layer of the Earth's atmosphere that is directly above the stratopause and directly below the mesopause.</span>
4 0
3 years ago
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Wine and cheese are complement goods because they are consumed together. What would we expect to happen to the equilibrium quant
notka56 [123]

Explanation:

As wine and cheese are consumed together,it means that they both are complimentary goods.

As per law of demand,the fall in price of a complimentary good would increase the demand and shift the demand curve of the other to the right.

The equilibrium quantity of cheese would increase and shift to right when the price of wine falls.

8 0
3 years ago
What net force is required to accelerate a car at a rate of 2m/s^2 if the car has a mass of 3,000kg? plus explanation with equat
Vinvika [58]

Answer:

<h2>6000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 3000 kg

acceleration = 2 m/s²

We have

force = 3000 × 2 = 6000

We have the final answer as

<h3>6000 N</h3>

Hope this helps you

3 0
2 years ago
A series RCL circuit is at resonance and contains a variable resistor that is set to 206Ω. The power dissipated in the circuit i
mash [69]

Answer:

Power dissipated in resistor 532 ohm is 0.503 watt

Explanation:

We have given in first case resistance R_1=206ohm

Power dissipated in this resistance is P_1=1.30watt

Power dissipated in the resistor is equal to P=\frac{v_{rms}}^2{R}

We have to find the power dissipated in the resistor is 1.30 watt

From the relation we can say that \frac{P_1}{P_2}=\frac{R_2}{R_1}

\frac{1.3}{P_2}=\frac{532}{206}

P_2=0.503watt

So power dissipated in resistor 532 ohm is 0.503 watt  

5 0
3 years ago
Consider two ideal gases, A and B, at the same temperature. The rms speed of the molecules of gas A is twice that ofgas B. How d
katen-ka-za [31]

The rms speed of the molecules of gas A is twice that of gas B. The molecular mass of A is one fourth to that of B.

Answer: Option B

<u>Explanation:</u>

Measuring the speed of particles at a given point in time results in a large distribution of values. Some molecules can move very slowly, others very fast, and because they are still moving in different directions, the speeds may be zero. (Velocity, vector quantity that corresponds to the speed and direction of the molecule.)

To correctly estimate the average velocity, you must take the squares of the mean velocity and take the square root of this value. This is known as the root mean square (rms) velocity and is shown as follows:

                 V_{r m s}=\sqrt{\frac{3 R T}{M}}

Where,

M – Gas’s molar mass

R – Molar mass constant

T – Temperature (in Kelvin)

Given data is rms speed for gas molecule A is twice that of gas molecule B. So,

                 \left(V_{r m s}\right)_{A}=2\left(V_{r m s}\right)_{B}

Therefore, equating the molecule’s rms speed formula for both A and B,

                  \sqrt{\frac{3 R T}{M_{A}}}=2(\sqrt{\frac{3 R T}{M_{B}}})

On squaring both sides, we get,

                 \frac{3 R T}{M_{A}}=4\left(\frac{3 R T}{M_{B}}\right)

By solving the above equations, we get,

                 M_{A}=\frac{M_{B}}{4}

8 0
3 years ago
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