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3 years ago

9

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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To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$u_{x}=u\cos\theta$

Put the value into the formula

$u_{x}=24\cos(31)$

$u_{x}=20.5\ m/s$

We need to calculate the vertical velocity

Using formula of vertical velocity

$u_{y}=u\sin\theta$

Put the value into the formula

$u_{y}=24\sin(31)$

$u_{y}=12.3\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{u_{x}}$

Put the value into the formula

$t=\dfrac{44}{20.5}$

$t=2.1\ sec$

We need to calculate the vertical height

Using equation of motion

$h=u_{y}t+\dfrac{1}{2}at^2$

Put the value into the formula

$h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2$

$h=4.2\ m$

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

$d=h-3.05$

Put the value of h

$d=4.2-3.05$

$d=1.15\ m$

Hence, The distance by the ball clear the crossbar is 1.15 m

7 0

3 years ago

What kind of friction exists between solid objects moving in water?

eduard

Fluid Friction exists when it is acted upon an object when in fluid.

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3 years ago

A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are mov

Kitty [74]

<span>net work = change in kinetic energy

for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction

10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²

now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)

10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8

they both will have the same velocity
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KE of block B=.5(1)(2.828²) = 4 J</span>

5 0

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Andrew [12]

Answer:

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Explanation:

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Levart [38]

Answer:

Explanation:

Given ,

dv / dt = k ( 160 - v )

dv / ( 160 - v ) = kdt

ln ( 160 - v ) = kt + c , where c is a constant

when t = 0 , v = 0

Putting the values , we have

c = ln 160

ln ( 160 - v ) = kt + ln 160

ln ( 160 - v / 160 ) = kt

(160 - v ) / 160 = $e^{kt}$

1 - v / 160 = $e^{kt }$

v / 160 = 1 - $e^{kt }$

v = 160 ( 1 - $e^{kt }$ )

differentiating ,

dv / dt = - 160k $e^{kt }$

acceleration a = - 160k $e^{kt }$

given when t = 0 , a = 280

280 = - 160 k

k = - 175

a = - 160 x - 175 $e^{kt}$

a = 28000 $e^{kt}$

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$e^{kt }$ = .00457

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3 years ago