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noname [10]
3 years ago
7

What components of a gas powered car are present in a hybrid? What components of an electric car or present in a hybrid? Interio

r answer in the space provided
Physics
1 answer:
rosijanka [135]3 years ago
4 0
I don't know how to anwser this but and gas hybrid uses and gas powered motor so tiny explosions inside of the cyiliders but the electric car uses and electric motor powered by a simple battery...I still don't know if this is correct....
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You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.
Colt1911 [192]

Answer:

Container A and C

Explanation:

ideal gas equation gives P=nRT/V

so at constant Temperature and pressure, P=n/T

Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

8 0
3 years ago
Acid rain is caused by human activity. Please select the best answer from the choices provided T F
olasank [31]

It is actually caused by the environment, so its false. :)

5 0
3 years ago
The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne
aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

ΔV_{H} =\frac{I}{nqt} B

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ\frac{V}{T}

b = —0.02  μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH =100X10^{-6} V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26}  } \\t=0.125mm

To learn more about the hall voltage , Visit: brainly.com/question/19130911

#SPJ4

8 0
2 years ago
Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.
Nataliya [291]

Answer:

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7 0
3 years ago
An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte
Klio2033 [76]

Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So,  V' < V

So, its speed will be less than V.

3 0
3 years ago
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