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3 years ago

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Two positive point charges that are equal in magnitude are fixed in place, one at x = 0.00 m and the other at x = 1.00 m, on the

x axis. A third positive point charge is placed at an equilibrium position. Where is the equilibrium position?

1 answer:

olga55 [171]3 years ago

6 0

Answer:

0.5 m

Explanation:

Two charges each of magnitude q

Let the third charge is Q is placed at a distance x from the origin so that the charge is in equilibrium.

The force on Q due to q at origin is balanced by the charge on Q due to the charge q placed at x = 1 m.

So,

$\frac{KQq}{x^{2}}=\frac{KQq}{\left ( 1-x \right )^{2}}$

1 - x = x

1 = 2x

x = 0.5 m

Thus, the third charge is placed at x = 0.5 m .

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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3 years ago

If the weight of 1kg is 10 N in water what is the density of the stone<br>

vekshin1

Answer:

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Explanation:

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3 years ago

Which of the following forms of energy must an object have if it is moving at a constant velocity?

lbvjy [14]

Answer:

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Explanation:

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3 years ago

10) If the mass 2m, the left mass

Romashka-Z-Leto [24]

Answer:

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$ .

Explanation:

Gravitational force between two objects of masses $m_{1}, m_{2}$ kept at a distance r is given by the formula

F = $\frac{-Gm_{1}m_{2} }{r^{2} }$

Here , $m_{1}$ = 2m

$m_{2}$ = $\frac{m}{2}$

Thus , F = $\frac{-G.2m.\frac{m}{2} }{r^{2} }$

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3 years ago

One horsepower (hp) is the amount of power required to lift a 75-kg mass a vertical distance of 1 m in 1 s. What is 2 hp equival

Vladimir [108]

Answer:

1470 W

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Energy/time

P = mgh/t ...................... Equation 1

Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.

Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.

Substitute into equation 1

P = (75×1×9.8)/1

P = 735 W.

From the above,

1 hp = 735 W

2 hp = (2×735) W

2 hp = 1470 W.

Hence 2 hp = 1470 W

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