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IRINA_888 [86]
2 years ago
4

Which of the following statements are true? Check all that apply. 1. In a circuit, current is delivered by the positive terminal

of a battery, and it is used up by the time it returns to the negative terminal of the battery. 2. When an electric field is applied to a conductor, the free electrons move only in the direction opposite the applied electric field. 3. By convention, the direction of a current is taken to be the direction of flow for negative charges. 4. In order to maintain a steady flow of current in a conductor, a steady force must be maintained on the mobile charges. 5. Current is the total amount of charge that passes through a conductor's full cross section at any point per unit of time. I've been reading this chapter in my book over and over and I still can't figure this out. I'd really appreciate some help! Thanks.
Physics
2 answers:
SOVA2 [1]2 years ago
7 0

Answer:

only statements 4 and 5 are correct.

4.In order to maintain a steady flow of current in a conductor, a steady force must be maintained on the mobile charges.

5. Current is the total amount of charge that passes through a conductor's full cross section at any point per unit of time.

Explanation:

<u>statement 4 is true:</u>

In order to maintain a steady flow of current in a conductor, a steady force must be maintained on the mobile charges.

<u>Reason:</u>

As we know there are three conditions for the current to flow as;

1. close loop

2. load or resistance

3. potential difference.

This potential difference is an example of a steady force which maintains a steady flow of current in a conductor.

<u>Statement 5 is true:</u>

Current is the total amount of charge that passes through a conductor's full cross section at any point per unit of time.

<u>Reason:</u>

As we know the definition of current is flow of charge per unit time i.e.

I=\frac{Q}{t}

where Q is the total charge and t is the time required for the charge to flow through a conductor.

alexandr402 [8]2 years ago
4 0

Answer:

4. In order to maintain a steady flow of current in a conductor, a steady force must be maintained on the mobile charges

5.Current is the total amount of charge that passes through a conductor's full cross section at any point per unit of time

Explanation:

5 is correct because current has the formula I=q/t which is the amount of charge per unit time .

4. Current flow has to be maintained with some amount of force due to its involvement with moving of electrons.

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max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
3. A snail crawls 5 inches in 15 minutes. What is its speed in in./min?
suter [353]

Answer:

3.0.33in/min...(c)

4.40m/min....(b)

5.10m/s....(a)

6.20mph...(b)

7.4m/s...(a)

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

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Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

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Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

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The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

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2 years ago
An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d
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Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

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2 years ago
A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
Semmy [17]

Answer:

The distance and height of the object  is 6 m and 2 m.

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Focal length = 0.25 m

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Using formula of magnification

m=-\dfrac{v}{u}

Put the value into the formula

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We need to calculate the height of the object

Using formula of magnification

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

6 0
2 years ago
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