Answer:
-22.3
Explanation:
Acellus
The net force on particle q2 is 21.7 N.
The given parameters:
Charge on particle 1, q1 = 8 μCCharge on particle 2, q2 = 3.5 μCCharge on particle 3, q3 = -2.5 μCDistance between particle 1 and particle 2 = 0.1 mDistance between particle 2 and particle 3, = 0.15 mThe net force on particle q2 is calculated as follows;
[tex]Q_2_{net} = q_{12} \ + \ q_{23}\\\\Q_2_{net} = \frac{kq_1q_2}{r_{12} ^2} \ + \ \frac{kq_2q_3}{r_{23}^2} \\\\Q_2_{net} = \frac{9\times 10^9 \times 8 \times 10^{-6} \times 3.5 \times 10^{-6} }{0.1^2} \ + \ \frac{9 \times 10^9 b\times 3.5 \times 10^{-6} \times (-2.5) \times 10^{-6} }{0.15^2} \\\\Q_2_{net} = 25.2 \ N \ - \ 3.5 \ N\\\\Q_2_{net} = 21.7 \ N[/tex]
Thus, the net force on particle q2 is 21.7 N.
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528 nm light passes through a single slit. the second (m=2) diffraction minimum occurs at an angle of 3.48 degrees. what is the width of the slit
Answer:
1.74
Explanation:
right on acellus
Answer:
1.74
Explanation:
acellus
IM BEGGING ILL GIVE BRAINLEST PLEASEEEEEEE HELP ME Distance and amount of charge are factors that determine which of the
following quantities?
A. Electrostatic force
B. Electric potential
C. Electric potential energy
D. All of the above
Explanation:
Distance and amount of charge are factors that determine which of the
following quantities?
D. All of the above
An auto, moving too fast on a horizontal stretch of mountain road, slides off the road,
falling into deep snow 43.9 m below the road and 87.7 m beyond the edge of the road. What was the acceleration 10m below the edge of the road?
Answer:
acceleration = - 9.8 m/s²
Explanation:
From the question, we can deduce that the only force that will be acting on the car is gravity and as such the acceleration during free fall will be equal to the acceleration due to gravity but will be negative since it is towards the ground.
Thus, acceleration = - 9.8 m/s²
1.00 x 100 kg of clear liquid (specific heat
capacity = 5.11 x 102 J/kg•°C) at a temperature
of 15.0°C gains 3.33 x 10 J of heat. What is the
final temperature of the liquid? (Assume the
melting point is less than 15.0°C and the boiling
point is greater than 62.0°C.)
Answer:
No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a ... to change 1 kg of liquid water at the normal boiling point (100ºC at atmospheric pressure) to steam (water vapor).
The melting point of lead is 327.3o C. Assume the final temperature of the system is T. Then the amount of energy released by the lead as it solidifies is. ΔQ = mleadLlead = 0.09 kg*(2.45*104 J/kg) = 2205 J
A star that is one of the coolest,
about 3,200°C, is going to be which
of the following colors?
Resourcos
A. greenish
Help
B. bluish
C. yellowish
D. reddish
Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 cm. If the current in one wire is 1.65 A and the current in the other wire is 3.25 A running in the opposite direction, determine the magnitude and direction of the force per unit length the wires exert on each other.
Answer:
The magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
Explanation:
Given;
distance between the parallel wires, r = 5.0 cm = 0.05 m
current in the first wire, I₁ = 1.65 A
current in the second wire, I₂ = 3.25 A
The magnitude of the force per unit length between the two wires is calculated as follows;
[tex]\frac{F}{l} =\frac{\mu_0 I_1 I_2}{2\pi r} \\\\\frac{F}{l} =\frac{4\pi \times 10^{-7} \times 1.65 \times 3.25}{2\pi \times 0.05} \\\\\frac{F}{l} = 2.145 \times 10^{-5} \ N/m[/tex]
Therefore, the magnitude of the force per unit length is 2.145 x 10⁻⁵ N/m and the direction of the force is outward or repulsive since the current in the two parallel wires are flowing in opposite direction.
A piece of irregularly shaped metal weighs 300N in air. When the metal is completely submerged in water, it weighs 232.5N. Find the volume and specific gravity of the metal.
Answer:
Volume of metal piece = 0.0069 m³ (Approx.)
Explanation:
Given:
Weight of metal in air = 300 N
Weight of metal in water = 232.5 N
Find:
Volume of metal piece
Specific gravity of metal
Computation:
We know that;
Density of water = 1,000 kg/m³
Buoyant force applied on metal piece = Weight of metal in air - Weight of metal in water
Buoyant force applied on metal piece = 300 N - 232.5 N
Buoyant force applied on metal piece = 67.5 N
Buoyant force = Volume of metal x Density of water x Gravitational force
67.5 = Volume of metal x 1,000 x 9.8
Volume of metal piece = 0.0069 m³ (Approx.)
Which of the following is NOT a geological event caused by the movement of tectonic plates?
A) Sand Dunes
B) Mountain forming
C) Earthquakes
D) Volcanoes
How do home computer scanners generally use electromagnetic waves?
O A. They change information received from sound waves into electromagnetic waves.
OB. They gather information from reflected and absorbed waves.
O C. They receive microwaves that are emitted by natural objects.
O D. They use X-rays that penetrate most objects. SUBMIT
Answer:
they gather information from reflected and absorbed waves.
Explanation:
A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the Horizontal. Find the distance the ball travels before it hits the ground.
Answer:
The horizontal distance traveled by the ball before it hits the ground is 64.42 m.
Explanation:
Given;
initial velocity, u = 27.0 m/s
angle of projection, θ = 30⁰
The horizontal distance traveled by the ball before it hits the ground is known as Range;
[tex]Range = \frac{u^2 sin(2\theta)}{g} \\\\Range = \frac{(27^2)sin(2 \times 30)}{9.8} \\\\Range = 64.42 \ m[/tex]
Therefore, the horizontal distance traveled by the ball before it hits the ground is 64.42 m.
a 0.1 kg object oscillates as a simple harmonic motion along x axis with a frequency f=3.185 hz. At a position x1 , the object has a kinetic energy of o.7 j and a potential energy 0.3 J.The amplitude of oscillation A is:
Answer:
The total energy must be .7 J + .3 J = 1 J for a particle at the endpoint or midpoint of motion.
Also, omega = (k / m)^1/2
f = omega / (2 * pi)
omega^2 = 4 pi^2 * f^2 = k / m
k = 4 * pi^2 * f^2 * m = 40.05
Max KE or PE = 1/2 k A^2
A^2 = 2 * E / k = 2 * 1 / k = .0499
A = .223 meters
which describes the attraction of one water molecule toward another
How does a battery generate electrical energy?
Since water is much denser than air, deep-sea divers experience a much higher ambient pressure underwater. Each 10 meters of depth underwater adds another 1 atm to the ambient pressure experienced by the diver. (Note: this is in addition to the 1 atm ambient pressure at the surface of the water!) What pressure, in psi, is experienced by a diver 50.0 meters below the surface of the water
Answer:
If you are at sea level, each square inch of your surface is subjected to a force of 14.6 pounds. The pressure increases about one atmosphere for every 10 meters of water depth. At a depth of 5,000 meters the pressure will be approximately 500 atmospheres or 500 times greater than the pressure at sea level.
Explanation:
At sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.
What is pressure?Pressure is defined as the force applied perpendicularly to an object's surface divided by the surface area over which it is applied.
Pressure is the physical amount of force exerted on a particular area.
Pressure can be expressed as
Pressure = Force / area
There are three types of pressure.
Absolute pressureGauge pressureDifferential pressureThus, at sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.
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3. Jose and Luke raced as they climbed to the top of the bleachers. Jose reached the top in 12.6 s and
Luke reached the top in 13.3 s. Jose and Luke have the same mass,
Who did more work?
How do you know?
Who had more power?
How do you know?
Answer:
jose
Explanation:
Jose run faster and did more work
because the time Jose took to accomplish the activity is less than Luke
both
both have same mass
The work done by both Jose and Luke is same because the change in potential energy is same.
The power of Jose is greater than than of Luke because Jose takes less time to do the same work.
What is power?Power is defined in physics as the amount of energy transferred or converted per unit time. The watt is the unit of power in the International System of Units, equal to one joule per second. Power is sometimes referred to as activity in older works. A scalar quantity is power.
Let the mass of each of them is m.
Hence, work done by each of them = mgh ( same for both of them)
As Jose reached the top in 12.6 s and Luke reached the top in 13.3 s.
Power of Jose = mgh/12.6
Power of Luck = mgh/13.3
Hence, Power of Jose is greater than Power of Luck.
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Finding Dimensional Formula For x=vt+1/2at is
Answer:
In the picture above.
Explanation:
I hope that it's a correct answer for dimensional analysis.
What do the different colors of stars tell us?
A. The size of the stars
B. The shape of the stars
C. The temperatures of the stars
Answer:
It would Be C The temperatures of the stars
Explanation:
Help answer question in licture
Answer:
D
Explanation:
D is high pitch c is low pitch
Answer:
D creo que esa es la nota mas alta, la segumda
Question 2 of 10
Which type(s) of radiation can damage living cells?
O A. Beta particles and gamma rays only
O B. Alpha particles, beta particles, and gamma rays
O C. Gamma rays only
D. Alpha particles and gamma rays only
The type of radiation which can damage the living cells are the X-rays, beta particles, and the gamma rays. Thus, the correct option is A.
What are damaging radiations?Radiations have been found to damage the DNA molecules present in the cells. High doses of the radiation damage the cell and these radiations could lead to Acute Radiation Syndrome (ARS) or Cutaneous Radiation Injuries (CRI). High doses of these radiation could also lead to formation of tumor and cancer later in the life.
Gamma rays are the most harmful external hazard or radiations among all the radiations and these result into acute effects. Beta particles are also harmful to the cell because these can partially penetrate the skin, causing beta burns in the skin. Alpha particles cannot penetrate the intact skin. Gamma and x-rays can pass through a person which results into damaging cells in their path.
Therefore, the correct option is A.
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I've got this question, which option is correct?
Answer:
option c is the correct one
The vector matrix [6 -2] is rotated at different angles. Match the angles of rotation with the vector matrices they produce.
Answer:
Hello your question is incomplete hence I will give you a general answer as regards rotation of vector matrix assuming angle of rotation = [tex]\frac{\pi }{4}[/tex]
answer :
angle of rotation = [tex]\frac{\pi }{4}[/tex]
vector matrix produced = attached below
Explanation:
lets assume the vector matrix [ 6, -2 ] is rotated clockwise by an angle of [tex]\frac{\pi }{4}[/tex]
The resultant matrix = attached below
Result of the rotation = attached below
attached below is the detailed solution
Answer: I got it right on Edmentum
Explanation: Answers attached below
sorry i was gone hope you dident mess me hers a question
Identify the organ that brings oxygen into the body.
brain
chest
heart
lungs
Answer:
lungs
Explanation:
lungs bring oxygen
Answer:
lungs
Explanation:
i took the quiz
PLEASE HELP ME PLEASE! BRAINLEST
Answer:
10kg
Explanation:
Weight is "how much does gravity drag this down".
Mass is "how much matter is there here".
The relation is:
[tex]F_g = mg[/tex]
where [tex]F_g[/tex] is the weight, [tex]m[/tex] is the mass and [tex]g[/tex] is the gravitational acceleration (roughly equal to 10N/kg on Earth).
From the task we know that:
[tex]F_g = 100N\\g = 10\frac{N}{kg}[/tex]
So let's input it into the relation:
[tex]100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg[/tex]
2.) Assume you decided to build a cabin off the electrical grid in a remote region of the United States, and your cabin would require 720 kWh per month to easily meet your needs. You decide to produce your electricity using photovoltaic cells and store it using a battery system so that it is available when you wish to use it. If every photovoltaic cell has the potential to produce 0.25 kilowatts (250 watts), and they produced electricity for 24 hours a day, how many photovoltaic cells would you need to meet your electricity needs for a month
Answer:
4 photovoltaic cells are required to meet monthly energy consumption of the cabin.
Explanation:
Let suppose that PV generation system produces energy at constant energy, the number of photovoltaic system to meet energy consumption of the cabin is:
[tex]n = \frac{E}{\dot E \cdot t}[/tex] (1)
Where:
[tex]E[/tex] - Energy consumption of the PV system, in kilowatt-hours.
[tex]\dot E[/tex] - Power generation of the PV cell, in kilowatts.
[tex]t[/tex] - Working times, in hours.
If we know that [tex]E = 720\,kWh[/tex], [tex]\dot E = 0.25\,kW[/tex] and [tex]t = 720\,h[/tex], then the number of photovoltai cells is:
[tex]n = \frac{E}{\dot E \cdot t}[/tex]
[tex]n = 4[/tex]
4 photovoltaic cells are required to meet monthly energy consumption of the cabin.
In Milgram's experiment:
A. Yale University would not allow the experiment to be done.
B. participants overwhelmingly did not go all the way to 450 volts.
C. the "learner" was working with Milgram.
D. the "teacher" was working with Milgram.
Answer:
B. The "Learner" was working with Milgram.
Explanation:
just took the test
give brainliest, please. :)
You have probably noticed that carrying a person in a pool of water is much easier than carrying a person through air. To understand why, find the buoyant force exerted by air and by water on the person. Assume the average volume of a person is 0.092 m3, and that the person is submerged in air and water respectively. (Give the magnitude.)
air: ________
water :_______
Answer:
The answer is "0.91238 and 744.8"
Explanation:
In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards:
[tex]f_b \to m \to mg =person \\\\F_B \ in\ air = v\ & air\ g \\\\[/tex]
[tex]=0.076 \times 1.225 \times 9.8 \\\\ =0.91238 \ N\\\\[/tex]
[tex]F_B \ in \ water = v \& water \ g \\\\[/tex]
[tex]=0.076 \times 1000 \times 9.8\\\\= 744.8 \ N\\[/tex]
What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)
Answer:
When we have a current I, we will have a magnetic field perpendicular to this current.
Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".
Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.
Of course, the intensity of the magnetic field is proportional to the current, given by:
B = (μ*I)/(2*π*r)
Where:
μ is a constant, I is the current and r is the distance between to the current.
Now remember that for a resistor:
R = ρ*L/A
R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.
If we increase the area of the wire (if we use a thicker wire).
And the relation between resistance and current is:
I = V/R
Where V is the voltaje.
Now, if we use a thicker wire, then the cross-section area of the wire increases.
Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.
And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.
If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.
A wheelbarrow full of bricks is lifted to the top of a wall. If the mass of the loaded wheelbarrow is 3000.0 g and the height of the wall is 0.45 meters, what is its GPE?
Answer:
the Gravitational potential energy is 13.23 J
Explanation:
The computation of the GPE is shown below:
GPE stands for Gravitational potential energy
The following formula should be used for the same
= mass × gravity × height
= 3000 g × 9.8m/sec^2 × 0.45 m
= 13.23 J
Hence, the Gravitational potential energy is 13.23 J
We simply applied the above formula so that we can easily determine the GPE
A negatively charged rod briefly touches a neutral metal ball. The metal ball will now be ____________
Answer:
let say it will be positive
Calculate the equivalent resistance for each of the following circuits.
Answer:
5. 60 Ω
6. 60 Ω
7. 10 Ω
8. 0.625 KΩ
Explanation:
5. Determination of the equivalent resistance.
Resistor 1 (R₁) = 10 Ω
Resistor 2 (R₂) = 20 Ω
Resistor 3 (R₃) = 30 Ω
Equivalent Resistance (R) =?
Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:
R = R₁ + R₂ + R₃
R = 10 + 20 + 30
R = 60 Ω
Thus, the equivalent resistance is 60 Ω
6. Determination of the equivalent resistance.
Resistor 1 (R₁) = 10 Ω
Resistor 2 (R₂) = 35 Ω
Resistor 3 (R₃) = 15 Ω
Equivalent Resistance (R) =?
Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:
R = R₁ + R₂ + R₃
R = 10 + 35 + 15
R = 60 Ω
Thus, the equivalent resistance is 60 Ω
7. Determination of the equivalent resistance.
Resistor 1 (R₁) = 6 Ω
Resistor 2 (R₂) = 4 Ω
Equivalent Resistance (R) =?
Since the resistors are arranged in series connection, the equivalent resistance can be obtained as follow:
R = R₁ + R₂
R = 6 + 4
R = 10 Ω
Thus, the equivalent resistance is 10 Ω
8. Determination of the equivalent resistance.
Resistor 1 (R₁) = 10 KΩ
Resistor 2 (R₂) = 2 KΩ
Resistor 3 (R₃) = 1 KΩ
Equivalent Resistance (R) =?
Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:
1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/10 + 1/2 + 1/1
Find the least common multiple (lcm) of 10, 2 and 1. The result is 10. Divide 10 by each of the denominator and multiply the result obtained by the numerator. This is illustrated below:
1/R = (1 + 5 + 10) / 10
1/R = 16/10
Invert
R = 10/16
R = 0.625 KΩ
Thus, the equivalent resistance is 0.625 KΩ.