Calculate the electric force between two point charges of -54.0 μC and +36.0 μC when they are 2.00 cm apart.

**note -- the metric prefix "micro" = μ = x10 - 6 **


Is this attractive or repulsive?


Can someone help me please? Thank you!!!!

Answers

Answer 1

Answer:

43740 N and attractive force

Explanation:

Given that,

Charge 1, q₁ = -54.0 μC

Charge 2, q₂ = +36.0 μC

The distance between charges, r = 2 cm = 0.02 m

We need to find the force between charges. The formula for the force between charges is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Put all the values,

[tex]F=9\times 10^9\times \dfrac{54\times 10^{-6}\times 36\times 10^{-6}}{(0.02)^2}\\\\F=43740\ N[/tex]

As the charges are opposite, the force between them is attractive. Hence, the required force is 43740 N.


Related Questions

The series circuit depicts three resistors connected to a voltage
source. The voltage source (AVtot) is a 110-V source and the resistor
values are 7.2 (R1), 6.2 A2 (R2) and 8.6 22 (R3).
b. Determine the current in the circuit.
A
c. Determine the voltage drops across each individual resistor.

Answers

Answer:

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V

Explain why a kangaroo can jump higher as it's speed increases

Answers

Answer:

See explanation

Explanation:

The ability of the Kangaroo to jump higher is due to the stretching of its tendons as its speed increases.

As the tendons are being stretched more and more, the elastic potential energy increases with each increasing stretch.

This elastic potential energy is translated into gravitational potential energy hence the Kangaroo jumps higher as it's speed increases.

The kangaroo can  jump higher as it's speed increases due to the ability of its tendons to stretch more and more.

The jumping of kangaroo can be related with the concept of spring force and spring potential energy. The spring force is given as,

F = kx

And spring elastic potential energy is,

[tex]U=\dfrac{1}{2}kx^{2}[/tex]

In both the cases, k is the spring constant and x is the stretching distance.

When kangaroo try to jump, the tendons provide extra flexibility to jump higher due to the more stretching. Here, the value of x increases. As value of x increases, ultimately more acceleration is provided to increase its speed.

As the tendons are being stretched more and more, the elastic potential energy increases with each increasing stretch.

This elastic potential energy is translated into gravitational potential energy hence the Kangaroo jumps higher as it's speed increases.

Thus, we can conclude that the kangaroo can  jump higher as it's speed increases due to the ability of its tendons to stretch more and more, such that more energy conversion of energy takes place.

Learn more about the elastic potential energy here:

https://brainly.com/question/25029446

If there are 1,600 meters in a mile, find the speed of sound in this experiment in units of "miles / hour"

Answers

Answer:

v = 771.75 mile / h

Explanation:

The speed of sound is v = 343 m / s  at T= 20C

the transformation ratios are

     1 mile = 1600 m

     1 h = 3600 s

Let's reduce

    v = 343 m / s (1 mile / 1600 m) (3600 s / 1 h)

    v = 771.75 mile / h

the answer is b :)

Samples of different materials, A and B, have the same mass, but the sample
of A is higher in density. Which statement could explain why this is so?
A. The particles that make up material B are more closely packed
together than the particles that make up material A.
B. The particles that make up material A have more mass than the
particles that make up material B.
C. The sample of material A has greater volume than the sample of
material B.
D. The particles that make up material B have more mass than the
particles that make up material A.

Answers

Answer:

Answer is letter B

Answer:  THE CORRECT ANSWER IS :  The partiles that make up material B are more closely packed together than the partiles that make up material A

Explanation:     I TOOK THE TEST    apex

How much energy is required to move an electron through a potential difference of
12 V?

1) 1.9 x 10^- 18 )
2) 7.5 x 10^- 18 )
3) 1.3 x 10^-20)
4) 7.5 x 10^18 J

Answers

I think it’s going to be the 2nd one

which one of the following is not a derived quantity a)speed b)velocity c)time d)force​

Answers

Answer:

c) time

Explanation:

time is a fundamental quantity from which other quantities are derived

A 20 ohm and 60 ohm light bulb are put in series together which one would be brighter?

Answers

Answer:

The 60 Ohm bulb would be brighter than the 20 Ohm bulb.

Explanation:

Hi there!

Ohm's Law: [tex]V=IR[/tex] where V is voltage, I is current and R is resistance

Power equation: [tex]P=VR[/tex]

The greater the power of a bulb, the brighter it is.

The bulbs are connected in series, meaning the the current travelling through each will be the same. Plug Ohm's law into the power equation so we just have I and R:

[tex]P=I^2R[/tex]

Because the current stays the same, the bulb with greater resistance will have greater power, hence being brighter. Therefore, the 60 Ohm bulb would be brighter than the 20 Ohm bulb.

I hope this helps!

Answer:

The 60 Ohm bulb would be brighter than the 20 Ohm bulb.

Explanation:

Hi there!

Ohm's Law: V=IRV=IR where V is voltage, I is current and R is resistance

Power equation: P=VRP=VR

The greater the power of a bulb, the brighter it is.

The bulbs are connected in series, meaning the the current travelling through each will be the same. Plug Ohm's law into the power equation so we just have I and R:

P=2RP=I

2

R

Because the current stays the same, the bulb with greater resistance will have greater power, hence being brighter. Therefore, the 60 Ohm bulb would be brighter than the 20 Ohm bulb.

I hope this helps!

Which of the following statements are true about covalent bonding between two atoms? Select all that apply.

A. Electrons are shared.
B. The electronegativities of the two atoms are close to each other.
C. The two atoms can be of the same element.
D. Electrons transfer from one atom to the other.

Answers

Answer:

A and C

Explanation:

Covalent bonding involves sharing by the atoms involved

Statements that can be considered as true statement about covalent bonding between two atoms are:

A. Electrons are shared.

B. The electronegativities of the two atoms are close to each other.

C. The two atoms can be of the same element.

Covalent bond can be regarded as chemical bond in which electrons pairs are been shared between atoms, these atoms can be of the same element.

These electron pairs are called pairs or bonding pairs.

Therefore, option A,B,C are correct.

Learn more at

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I WILL MARK YOU THE BRAINLIEST LINKS WILL BE REPORTED
put the correct words with the correct pictures

Answers

1, 2, 4, 5 is examples of mass everything else is weight!

A uniform plank 4 m long is pivoted about its center to an angle of 30° with the vertical. What magnitude of force
must be applied 0.5 m from the pivot to maintain equilibrium with a 4 kg mass hanging from one end?
A. 80 N
B. 20 N
C. 40 N
D. 60 N
E. 100 N

Answers

Answer:

A. 80 N

Explanation:

Since the uniform plank is pivoted at its center at an angle of 30 to the vertical, then it is 90° - 30° = 60° to the horizontal.

Let L be the length of the plank = 4 m, the perpendicular distance of the 4 kg mass form the pivot point is Lcos60°/2. Then the moment of the 4 kg mass about its pivot point is mgLcos60/2 where m = mass = 4 kg and g = acceleration due to gravity = 10 m/s²

Let F be the force applied at 0.5 m from the pivot point. The perpendicular distance of F from the pivot point is 0.5cos60°. Thus the moment of this force about the pivot point is F × 0.5cos60° = 0.5Fcos60°

Since both moments are the same for equilibrium,

mgLcos60/2 = 0.5Fcos60°

F = mgL/(2 × 0.5 m)

F = 4 kg × 10 m/s² × 4 m/1 m

F = 80 kgm/s²

F = 80 N

Weathering of rocks can occur in many ways. In the western United States, strong winds can erode huge rock formations by blowing millions of tiny grains of sand at these rocks. Which term accurately describes this type of weathering?

Answers

Answer: physical or mechanical weathering

Explanation:

Mechanical weathering which is also referred to as the physical weathering occurs when a rock is broken down into smaller pieces. In this case, there will be a physical change of the rock but its composition will not change.

Some examples include ice freezing and expansion of the cracks in the rock, Smstrong winds that carrycpieces of sand which then sandblast surfaces, moving water which causes abrasion etc.

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resisting motion, including friction and air resistance, total 432.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 241 kg

Answers

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

[tex]a=3 \ m/s^2[/tex]

Forces resisting motion,

[tex]F_f=432 \ N[/tex]

Mass,

[tex]m = 241 \ kg[/tex]

By using Newton's second law, we get

⇒ [tex]F-F_f=ma[/tex]

Or,

⇒         [tex]F=ma+F_f[/tex]

By putting the values, we get

⇒             [tex]=(3\times 241)+432[/tex]

⇒             [tex]=723+432[/tex]

⇒             [tex]=1155 \ N[/tex]

A 32.0 kg cart is at rest on a frictionless, horizontal surface. The cart experiences an unbalaced force, causing the cart to accelarate at a rate of +5 m/s2.
Calculate the time it takes for the cart to travel 200 meters.

Answers

Answer:

t = 8.94 s

Explanation:

We will use the second equation of motion to calculate the time taken by the cart:

[tex]s = v_it + \frac{1}{2} at^2[/tex]

where,

s = distance = 200 m

vi = initial speed = 0 m/s

t = time taken = ?

a = acceleration = 5 m/s²

Therefore,

[tex]200\ m = (0\ m/s)(t)+\frac{1}{2}(5\ m/s^2)t^2\\\\t^2 = \frac{(200\ m)(2)}{5\ m/s^2}\\\\t = \sqrt{80\ s^s}[/tex]

t = 8.94 s

For general projectile motion with no air resistance, the vertical component of a projectile's acceleration For general projectile motion with no air resistance, the vertical component of a projectile's acceleration remains a non-zero constant. continuously decreases. first decreases and then increases. is always zero. continuously increases.

Answers

Answer:

Rmains constant

Explanation:

The equation of the trajectory of a projectile motion is presented as follows;

[tex]Y = x \cdot tan \theta -\dfrac{g \cdot x^2}{2 \cdot u^2 \cdot cos^2 \theta}[/tex]

The vertical componet of the prjectile motion is

y = (u·sinθ)·t - g·t²/2

Where;

θ = The angle with which the projectile is launched

x = The horizontal distance

u = The initial velocity of the projectile

g = The acceleration due to gravity = Constant

t = The time of motion

The acceleration acting on the projectile is the 'g' which is the constant acceleration due to gravity

Therefore, for general projectile motion with no air resistance, the vertical component of the projectile acceleration remains constant

what 3 words or 3 phrases that would help you remember the steps to Covalent and Ionic bonding?

Answers

Answer:

non metals and non metals=covalent bond

metals and non metals =ionic bond

transfer all the valence electron to form ionic bond

share valence electron to form covalent bond

A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height it was released from.

Answers

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

[tex]mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}[/tex]

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}[/tex]

h = 0.46 m

why we need system of measurement​

Answers

Answer:

Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and the chemical industry, it is also essential in farming, engineering, construction, manufacturing, commerce, and numerous other occupations and activities.

Explanation:

A truck starts from rest with an acceleration of 0.3 m/ S^2 find its speed in km/h when it has moves through 150 m​

Answers

Answer:

v = 34.128 km/hr

Explanation:

Given that,

The initial speed of a truck, u = 0

Acceleration of the truck, a = 0.3 m/s²

Distance moved, d = 150 m

Let the final speed of the truck is v. Using third equation of motion i.e.

[tex]v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad}[/tex]

Put all the values,

[tex]v=\sqrt{0^2+2\times 0.3\times 150}\\\\v=9.48\ m/s[/tex]

or

v = 34.128 km/h

So, the final speed of the truck is equal to 34.128 km/h.

A 8.25-N force is applied to a 2.50 kg object to accelerate it rightwards. The object
encounters 2.19 N of friction. Determine the acceleration of the object. (Neglect air
resistance.)

Answers

Answer:

Explanation:

Use F - f = ma since we are on a flat surface. We are given the applied Force as 8.25, the frictional force as 2.19, and a mass of 2.50. Filling in:

8.25 - 2.19 = 2.50a. Doing the subtraction on the left gives us:

6.06 = 2.50a and we divide to 3 sig figs to get

a = 2.42 m/s/s

The magnetic field 0.100 m from a
wire is 4.20 x 10-5 T. What is the field
0.200 m from the wire (twice as far)?
[?] *10^[?]T

Answers

Answer:

Your answer is given below:

Explanation:

A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approximate acceleration of the car?

Answers

Answer:

Explanation:

The equation for acceleration is

[tex]a=\frac{v_f-v_0}{t}[/tex]. We have the initial velocity since we are told that the car started from rest. What we don't have is the final velocity. But we can find it because we were told that the car undergoes uniform acceleration, so the equation

d = rt will give us that final velocity.

120 = r(6.32) so the velocity or rate of the car is

r = 19 m/s. Plug that in to find the acceleration of the car:

[tex]a=\frac{19-0}{6.32}[/tex] so

a = 3.0 m/s/s

What happens to the mass number of an atom when the
number of neutrons in the nucleus of that atom increases?

Answers

pretty sure it increases

Transformar las siguientes unidades al Sistema Internacional: 30 km/h ; 37 Dm ; 750 g ; 4x10-6 km2 ; 7500 cm ; 600000 cm2 ; 520700000 mm3 ; 3,4 años.

Answers

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

A magnetic compass is placed near an insulated copper wire. When the wire is connected to a battery and a current is created, the compass needle moves and changes its position. Which is the best explanation for the production of a force that causes the needle to move?

Answers

Explanation:

When the wire is connected to a battery, the compass needle moves and changes its position. This happens because the needle magnetizes the copper wire, thus,  creating a force.

While the current in the wire produces a magnetic field and exerts a force on the needle. The insulation on the wire becomes energized and exerts a force on the needle. Hence, the compass needle moves and changes its position.

8× +5+9×+3 how can I solve this​

Answers

Answer:

collect like terms then add

=>8x+9x+5+3

=>8x+9x+5+3=>17x+8

if molecule gains heat energy it converts liquid to gas​

Answers

Answer:

Actually, the substance is the one that changes from liquid to gas.

The molecule just gains energy which overcomes the attraction force between it and neighboring molecule.

A rocket blasts off from Earth with an initial
acceleration of 18.2 m/s2. A 75.0 kg astronaut
is inside. What is the normal force of the
astronaut's chair on the astronaut?
S
[?]N

Answers

Explanation:

apparent apparent weight is how hard astronaut pushes against it which is the normal force

n m a y + mgm open parenthesis y + and close parenthesis and = m open parenthesis a y + and so i need the acceleration of the astronaut since initial velocity is 0 and yby over l 45 m per second over 15 x 30.0 m m per second squared

What happens to the wavelength of a wave if the frequency quadruples, but the wave is in the same medium?

A. The wavelength will go down to half the original amount.

B. The wavelength will go down to one-fourth of the original amount.

C. The wavelength will be quadruple.

D. The wavelength will double.

Answers

Answer:

I think C? I'm not sure totally though...

Explanation:

A 0.2 kg hockey puck is sliding along the ice with an initial velocity of -10 m/s when a player strikes with his stick, causing it to reverse it's direction and giving it a velocity of +25 m/s. The impulse the stick applies to the puck is most nearly

Answers

I = 7 N-s

Explanation:

Impulse is defined as

I = ∆p = m(vf - vi)

We are given

m = 0.2 kg

vi = -10 m/s (to the left)

vf = +25 m/s (to the right)

Therefore, the impulse imparted on the puck is

I = (0.2 kg)[25 m/s - (-10 m/s)]

= (0.2 kg)(35 m/s)

= 7 N-s

3. How do energy transformations, energy transfers, and conservation of energy allows you to track how energy moves through in a system? ​

Answers

Answer:

The law of conservation of energy states that when one form of energy is transformed to another, no energy is destroyed in the process. According to the law of conservation of energy, energy cannot be created or destroyed. So the total amount of energy is the same before and after any transformation

Explanation:

The energy in a system always transforms from one form to another without any significant loss or addition. It simply changes from one form to another with the effect of numerous external influences.

What is Energy transformation?

Energy transformation may be defined as the process through which energy can convert from one form to another. It also illustrates the migration of energy from one place to another due to physical factors.

According to the law of conservation of energy, in a closed system energy can neither be created nor destroyed. The total amount of energy remains the same before and after the process of transformation.

Therefore, the energy in a system always transforms from one form to another without any significant loss or addition.

To learn more about Energy transformations, refer to the link:

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