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andreev551 [17]

3 years ago

12

How strong is the repulsive force exerted on two point charges that each carry 1.0 E-6 C of negative charge and are 0.30 meters

apart? Coulomb’s Constant is 9.0 E9 N*m^2/C^2 Remember to identity all of your data, write the equation, and show your work.

2 answers:

Olin [163]3 years ago

8 0

Answer

Force between two charged bodies q₁ and q₂ is given by the formula below.

F = Kq₁q₂/r²

Where k = constant

r = distance between them

F = (9.0 E9 × 1.0 E-6 × 1.0 E-6)/0.30²

= 9.0E-3/0.30²

= 9.0E-3/0.09

= 0.1 N

frutty [35]3 years ago

3 0

Coulomb's law states that the force between two charged particles is directly proportional to the multiplication of the two charges and inverse with the square of the distance between them.

<u>Mathematically,</u>

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

where: F is the force in (N), $q_{1},q_{2}$ are the charge of the two particles in (C), r is the distance between them and k is Coulomb's constant

<u>Thus</u>

$F= 9.0*10^9* \frac{1.0*10^-6*1.0*10^-6}{0.3^2} =0.1N$

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How old is a bone if it still has 50% of its carbon-14 content?

Degger [83]

Answer:

5730 years

Explanation:

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The half-life equation is:

A = A₀ (½)^(t / T)

where A is the remaining amount,

A₀ is the initial amount,

t is time,

and T is the half life.

In this case, A = ½ A₀ and T = 5730.

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3 years ago

an object is 40M below the surface of sea water of density 1020 kg/m3 calculate the pressure difference

olga nikolaevna [1]

Answer:

P = 400248 [Pa]

Explanation:

We can calculate the hydrostatic pressure exerted by a column of liquid by means of the following equation.

$P=Ro*g*h$

where:

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Therefore:

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3 years ago

A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara

saw5 [17]

Answer:

(a) $\alpha = 53.73 m/s^2$

(b) I =428 $kgm^2$

(c) $\tau = 428 \times 53.73 = 22996 .44Nm$

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

$\omega = \frac{v}{l}$

$\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s$

for angular acceleration, use the following equation.

$\omega _{f}^2 = \omega_{i}^2+2\alpha\theta$

since $\omega _{i} = 0$

here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians

so for one quarter $\theta = \frac{\pi }{2}$

$(12.99)^2 = 2\alpha(\frac{\pi }{2})$

on solving

$\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2$

(b)

For the catapult,

moment of inertia

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2$

For the ball,

$I = MR^2$

$I = 18.2 \times 14.51$

$I = 264 kgm^2$

so total moment of inertia = 428 $kgm^2$

(c)

$\tau = I\alpha$

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What formula can be used to determine the velocity of an object in free fall?

PtichkaEL [24]

Assuming that you're given either an initial or final velocity, you can use the following equation and solve for the initial or final velocity.

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g = 9.81 m/s²
(y - y₀) = the change in the distance along the y-axis.

You'll need also determine the positive and negative of your y-axis for your final solution because velocity can be positive or negative based on direction. Lastly, don't forget to square root both sides of your equation for your velocity.

I hope this helps.

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