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grigory [225]
3 years ago
13

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Determine

the force between the two balloons.
Physics
1 answer:
andrew-mc [135]3 years ago
4 0

F = 0.06N. Since the charges has different signs the force of atraction between them is 0.06N.

In order to solve this exercise we have to use Coulomb's Law equation which says that the magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. Given by the equation:

F = k\frac{|q_{1}q_{2}|}{d^{2}}. Where k is the Coulomb's Constant k = 9x10^{9}\frac{Nm^{2} }{C^{2} }, q1 and q2 are the charges value in Coulomb (C), and d is the distance between charges in meters (m).

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Calculate the force between the two ballons.

F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|(3.37x10^{-6}C)(-8.21x10^{-6}C)|}{(2.00m)^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|-2.77x10^{-11}C^{2} )|}{4.00m^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} }(6.92x10^{-12})\frac{C^{2} }{m^{2}}\\F = 0.06N

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Convert 75.0 degrees Fahrenheit into Celsius
nikdorinn [45]

Answer:

23.889 Celcius

Explanation:

(75°F − 32) × 5/9 = 23.889°C

3 0
3 years ago
Read 2 more answers
Aluminum oxide can be produced during rocket launches. Show that the sum of positive and negative charges in a unit of Al2O3 equ
Triss [41]

Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.

Al203 has 2 Al atoms and 3 O atoms.

Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)

= 2(3) + 3(-2)

= 6 - 6

= 0

Explanation:

Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.

Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.

4 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0
2 years ago
Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec
nadezda [96]

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\

On substituting the above values we get the required acceleration

Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}

Therefore,the acceleration of the train is 5 m/s².

4 0
3 years ago
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