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geniusboy [140]
3 years ago
5

Which type of resistor is commonly used in automotive circuits? a. Fixed value b. Stepped c. Variable d. All of the choices

Physics
1 answer:
soldi70 [24.7K]3 years ago
7 0

Answer:d-All of the above

Explanation:

A resistor is a device used in electrical circuits that offers resistance to current flow.In automotive circuit commonly used resistor are

  1. Fixed value
  2. Stepped
  3. Variable

Fixed value:A fixed value electrical resistance is one whose value is resistance value is fixed and does not change in circuit.

Stepped resistance:Stepped resistor also called as tapped resistors provide two or more set resistor values. In a circuit, the various resistances are attached to different terminals. When the switch is shifted, different resistance values are put in the circuit.

Variable resistance:The variable resistor gives more control over the current flow by increasing the resistance. When the resistance of the variable resistor increases, the amount of current allowed to flow in the circuit decreases.              

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1) What do (x) and (y) symbolize?
MArishka [77]

Answer:

x is vertical and y is horizontal

Explanation:

7 0
3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
How do you give an example for finding net force?
Crazy boy [7]
--  6 people all trying to push a car out of snow

--  a Tug-o-War with 30 people of different sizes pulling on each end of the rope

--  you and your sister both pulling on the same doll (or Transformer)

--  lifting a book up from the table to a high shelf
    taking a book down from a high shelf to the table
    (one force is you; another force is gravity)

--  grabbing your big dog by his collar and trying to bring him inside

--  three people at the table all grab the ketchup bottle at the same time
7 0
3 years ago
imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north
mylen [45]
You will have to fly around the whole earth to get to your landing station
8 0
3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
4 years ago
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