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2 years ago

9

Tech A says that a vacuum modulator converts manifold vacuum into an engine load signal. Tech B says that the manual valve is mo

ved by governor pressure to force a shift. Who is correct?
a. tech A
b. tech B
c. both A and B
d. neither A nor B

1 answer:

alexandr1967 [171]2 years ago

8 0

Answer:

The correct option for the answer is A.) Tech A

Explanation:

Tech A is correct

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What is the slope of the line?

coldgirl [10]

(30, 5)
(10, 1)

change of y / change of x
= (30 - 10) / (5 - 1)
= 20 /4
= 5

6 0

2 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat

jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{24}$

time = 2 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{72}$

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0

3 years ago

A car is traveling at a constant speed on the highway. Its tires have a diameter of 65.0 cm and are rolling without sliding or s

lesya692 [45]

Answer:

Explanation:

The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16

4 0

2 years ago

What happens to a liquid beyond its certain temperature

Natalka [10]

Answer: what do you mean ''beyond certain temperature'', Do you mean ask in cooling the liquid down or heating the liquid up?

Explanation:

7 0

2 years ago