Ask question

Ask question

All categories

2 years ago

10

Technician A says that the device that controls shift points on an automatic transmission is the valve body. Technician B says t

hat the device that controls shift points on an automatic transmission is the computer. Who is correct?
A) Technician A only
B) Technician B only
C) Both A and B
D) Neither A nor B

1 answer:

alexira [117]2 years ago

3 0

Answer: Technician B only

Explanation:

What actually controls the shifting in a automatic transmission is the vehicle's internal computer that helps shift the gears automatically. This includes a Control Module that adjusts the current provided to solenoids inside to control the gears The hydraulic system is used to select gears dependent on the transmission fluid's pressure applied.

You might be interested in

** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

$V_p = 350,000 Volts$

$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

6 0

2 years ago

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m

Angelina_Jolie [31]

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

$U(x)=2x^3-15x^2+36x-23$

and we know force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

$F=-(2\times 3x^2-15\times 2x+36)$

For Force to be zero F=0

$\Rightarrow 6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow x^2-2x-3x+6=0$

$\Rightarrow (x-2)(x-3)=0$

Therefore at x=2 and x=3 Force on particle is zero.

8 0

2 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago

A train is moving at 15ms-1. It slows down to 10.5ms over a time of 4s. Calculate the distance it travels before stopping if acc

Nostrana [21]

Answer:

3WGGRGWERGRG

Explanation:

GERAGAETGAERR GHERUERHGRRGF;SBDF;JKSRDMFNSDFLDGGD;GDVF

4 0

2 years ago

Bone with a latticework structure is called

malfutka [58]

Spongy or the cancellous bone is the bone with lattice work structure

7 0

2 years ago