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3 years ago

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

Fixed value
Stepped
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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you are given two circuits with two batteries of emf e and internal resistance r1 each. circuit a has the batteries connected in

jeka57 [31]

Answer:

In which direction does the current in circuit A flow?

counterclockwise

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Explanation:

5 0

3 years ago

A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for

inysia [295]

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

It starts from rest.
From t = 0 s to t = 4.0 s it is accelerated with a constant force.
From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

$a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s} = 0 m/s^{2}$

Learn more: brainly.com/question/16274121

3 0

2 years ago

Read 2 more answers

Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago

A ball is kicked from a height of 20 meters above the ground. If the initial velocity is 10 m/s, how long is the ball in flight

nasty-shy [4]

S=20 m
v=10 m/s
t=s/v
= 20/10
= 2 s.

8 0

3 years ago

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3

Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

w = 2π f = 2π / T

2π / T = √ k / m

k₀ / m₀ = (2π / T)²

k₀ / m₀ = (2π / 3.0)²

k₀ / m₀ = 4.3865

The period on the new planet is

2π / T = √ k / m

T = 2π √ m / k

In this case the amounts are

m = 6 m₀

k = 10 k₀

We replace

T = 2π√6m₀ / 10k₀

T = 2π √6/10 √m₀ / k₀

T = 2π √ 0.6 √1 / 4.3865

T = 3.23 s

5 0

3 years ago