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3 years ago

Which of the following scenarios will cause the current to stop flowing within a circuit?

2 answers:

8 0

A. The switch within the closed circuit is opened.

If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.

n200080 [17]3 years ago

6 0

On page 39 it explains that conducting materials. If some part of the external circuit is composed of an insulating material— for example, if there’s an air gap in the circuit—it’s an open circuit. An electric current will flow through a circuit only if the circuit is closed. A common method of controlling the flow of current through a circuit is to include a device called a switch. Current will flow through the circuit when the switch is closed and will stop flowing when the switch is open.

So your answer is A. The switch within the closed circuit is opened

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7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist

luda_lava [24]

Answer: 5.05 m

Explanation:

4 0

3 years ago

Read 2 more answers

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

Given:

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

3 years ago

Answer ?

morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

5 0

2 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

Help me find the acceleration

ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the timeless suvat equation is likely what you need for this question.

In the timeless suvat equation,

$a = \dfrac{v^2 - u^2}{2\; x}$

where

$a$ is the acceleration of the car;
$v$ is the final velocity of the car;
$u$ is the initial velocity of the car; and
$x$ is the displacement of the car.

Note that v and u are velocities. Make sure that you include their signs in the calculation.

In this question,

$a$ is the unknown;
$v = -10.9 \; \text{m} \cdot \text{s}^{-2}$ ;
$u = -27.7 \; \text{m} \cdot \text{s}^{-2}$ ; and
$x = - 105 \; \text{m}$ .

Apply the timeless suvat equation:

$a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}$ .

The value of $a$ is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0

3 years ago