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Agata [3.3K]
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
Physics
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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How does a decrease in thermal energy affect the particle motion in each state of matter
Valentin [98]
The thermal energy is proportional to the movement of the particles in every state.

Decreasing the thermal energy will decrease the movement.
3 0
3 years ago
A system releases 255 cal of heat to the surroundings while delivering 428 cal of work. what is the change in internal energy of
vichka [17]

Answer:-683 cal

Explanation:

Given

Heat released by system Q=-255 cal

as heat released is taken as negative and vice-versa

Work done by system W=428 cal

From First law of thermodynamics

Q=\Delta +W , where \Delta U=change in internal Energy

\Delta U=-255-428=-683

7 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

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