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VashaNatasha [74]

3 years ago

5

Each circled letter in the circuit diagram represents a meter that is used to measure a quantity in the circuit. Which meters ar

e ammeters?
A. B and D

B. A and C

C. A and B

D. C and D

2 answers:

Tamiku [17]3 years ago

5 0

Answer:

The correct option is B): meters A and C.

Explanation:

rosijanka [135]3 years ago

3 0

<span>The correct option is B): meters A and C. In fact, both A and C are ammeters. An ammeter is a device used to measure the current in a branch of a circuit. In order to work properly, an ammeter should be connected in series to a resistor, in order to have the same amount of current that passes through the resistor. In this circuit, we see that devices A and C are both connected in series to a resistor, so they are ammeters. Instead, devices B and D are connected in parallel, so they do not measure the current but the voltage (in fact, they are called voltmeter).</span>

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Match the variables to its definition

MA_775_DIABLO [31]

Answer:

See connections below

Explanation:

1 $\Rightarrow$ b

2 $\Rightarrow$ a

3 $\Rightarrow$ d

4 $\Rightarrow$ i

5 $\Rightarrow$ g

6 $\Rightarrow$ h

7 $\Rightarrow$ c

8 $\Rightarrow$ e

9 $\Rightarrow$ f

4 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago

What is the name for family labeled #4 (Yellow)?

goldfiish [28.3K]

Answer:

transition metals im sorry if this was too late

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3 years ago

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$

8 0

3 years ago