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2 years ago

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What happens to the charge on the conductive sphere when it is connected to a source of charge such as the electrostatic voltage

source?

1 answer:

hichkok12 [17]2 years ago

8 0

Answer and Explanation:

The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.

Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.

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A person riding in a hot air air balloon basket ascends to 1000 ft above the ground. What happened to the atmospheric pressure a

kherson [118]

C) it decreased
because atmospheric pressure decreases as we move up.

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3 years ago

the candle had 350 J of energy in the chemical store. As it burned, 250 J was transferred by lighting to the surrounding

Elan Coil [88]

Answer:

Efficiency = 71%

Explanation:

Given the following data:

Output energy = 250 Joules

Input energy = 350 Joules

To find efficiency;

$Efficiency = \frac {Out-put \; energy}{In-put \; energy} * 100$

Substituting into the equation, we have;

$Efficiency = \frac {250}{350} * 100$

$Efficiency = 0.7143 * 100$

Efficiency = 71.43 ≈ 71%

Therefore, the efficiency of the candle is 71 percent.

4 0

3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d)

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

So

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

kkurt [141]

The answer is Anguer...

<em>Hope </em><em>it </em><em>helps.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>

3 0

2 years ago

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

<em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>

<em>The relationship is no longer linear and v will be more for the same value of x</em>

<em>The relationship is still linear, with lesser value of v</em>

<em>The relationship is still linear, with higher value of v</em>

<em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

6 0

2 years ago