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3 years ago

Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.

Physics

1 answer:

n200080 [17]3 years ago

8 0

Answer:

The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

F = k q₁q₂ /r²

-Substitution

F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

F = 3.512 x 10⁻⁷ / 0.381

-Result

F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N

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Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
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r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

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2 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

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Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

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$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

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$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

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3 years ago

A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s. During this time, the rol

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The initial velocity is 0.65 m/s

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The motion of the roller coaster is a uniformly accelerated motion, so we can solve the problem by using the following suvat equation:

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s is the displacement

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v = 35 m/s

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3 years ago

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Answer:

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Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

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3 years ago

The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect

Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

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T = 0.44K

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3 years ago