All categories

2 years ago

What type of configuration is used in the HEVs available in the US today?

Physics

1 answer:

Citrus2011 [14]2 years ago

6 0

Answer:

Currently in the united states using parallel system

Explanation:

because you can walk with the twomodes with internal combustion engine or running on electric power.

You might be interested in

The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section $=3.7\times 3.7 cm^2=13.69 cm^2$

N=no of turns=61

Resistance= $R=8 \Omega$

current (i)=0.04 A

emf induced $=i\times R=0.04\times 8=0.32 V$

emf induced $e=NA\frac{\mathrm{d} B}{\mathrm{d} t}$

$0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}$

$\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s$

8 0

3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago

A ball rolls from rest at the top of a 15.5 m hill, downward to the right and back upward onto a 8.25 m hill. How fast is the ba

lana66690 [7]

Answer:

Explanation:

6 0

3 years ago

Which statement best explains why an object

slavikrds [6]

D, Mercury as a weaker gravitational pull! Due to mercury being farther from the sun and it being a smaller planet it has a weaker pull

4 0

3 years ago

Read 2 more answers

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of

Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0

3 years ago