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3 years ago

What happens when an electrically charged balloon is placed close to an uncharged soda can?

Physics

2 answers:

Gemiola [76]3 years ago

6 0

Answer:

The surface closest to the balloon becomes positively charged, while the far side of the can becomes negatively charged. This results in an attractive force that draws the can to the balloon.

Explanation:

Artyom0805 [142]3 years ago

5 0

Answer:

The balloon is negatively charged.

Explanation:

hope it helps

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6. Decelerating a plane at a uniform rate of -8 m/s2, a pilot stops the plane in 484 m. How

uysha [10]

Answer: 88 m/s

Explanation:

If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

Where:

$V_{f}=0$ Is the final velocity of the plane (we know it is zero because we are told the pilot stops the plane at a specific distance)

$V_{o}$ Is the initial velocity of the plane

$a=-8m/s^{2}$ is the constant acceleration of the plane

$d=484m$ is the distance at which the plane stops

Isolating $V_{o}$ from (1):

$V_{o}=\sqrt{-2ad}$ (2)

$V_{o}=\sqrt{-2(-8m/s^{2})(484m)}$ (3)

Finally:

$V_{o}=88m/s$ This is the veocity the plane had before braking began

8 0

3 years ago

Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular

Agata [3.3K]

Answer:

The angular displacement is $\theta = 29.6 \ rad$

Explanation:

From the question we are told that

The initial angular speed is $w = 5.35 \ rad/s$

The angular acceleration is $\alpha = 0.331 rad /s^2$

The time take is $t = 4.81 \ s$

Generally the angular displacement is mathematically represented as

$\theta = w * t + \frac{1}{2} \alpha * t^2$

substituting values

$\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2$

$\theta = 29.6 \ rad$

3 0

3 years ago

A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge

Simora [160]

Answer:

(b) the point charge is moved outside the sphere

Explanation:

Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.

$\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.

If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.

Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.

7 0

4 years ago

The acceleration due to gravity on Mars is 3.60 m/s2. if this is measured at a distance of 3.4 x 10^6 m from the center of the p

densk [106]

As we know that acceleration due to gravity is given by

$g = \frac{GM}{r^2}$