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3 years ago

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Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC.

Which one of the following statements is true about the magnitudes of the electric forces on these beads

2 answers:

steposvetlana [31]3 years ago

3 0

The electric forces on both beads are equal.

Leni [432]3 years ago

3 0

Answer:

The charges repel with different forces

Explanation:

The charges repel with different magnitudes. This can be seen from the equation:

$F = k\frac{q_{1}q_{2} }{r^{2} }$

where q₁ and q₂ are the charges of the particles

k is Coulomb constant = 8.99 × 10⁹ N $m^{2} C^{-2}$

r distance between the charges

So the force will be $F = 8.99 * 10^{9} \frac{10^{-6}*1^{-6} }{(0.25^{2} )}$

= 143 840 N

This is a large force, which shows that the forces repel each other.

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True [87]

Answer: Student 2

Explanation: Iron nail and a paperclip are conductors because they are made of metal. A rock, rubber band, and wooden stick are insulators because they cannot conduct electricity.

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2 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)

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3 years ago

Two boats start together and race across a 48-km-wide lake and back. Boat A goes across at 48 km/h and returns at 48 km/h. Boat

nika2105 [10]

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

$d = 48 km$

now the time taken by the boat to move to and fro is given as

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

$t = \frac{48 + 48}{48}$

$t = 2 hrs$

Time taken by Boat B to cover the distance

$t = \frac{48}{24} + \frac{48}{72}$

$t = 2.66 h$

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

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3 years ago

If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil

AfilCa [17]

Answer:

Explanation:

Let the volume of air be V. at atmospheric pressure, that is 10⁵ Pa

At 20 m below surface pressure will be

atmospheric pressure + hdg

10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

At this pressure volume V becomes V/ 2.96

This volume will last 1/2.96 times time that is 60/2.96 = 20.27 minutes.

8 0

3 years ago

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3 years ago