Answers
Answer:
240 v
Explanation:
It will be twice as much (since it has twice as many turns) = 240 v
Related Questions
The Earth wire only carries a current if there is a _____ in the
circuit. What one word completes the sentence?
Answers
Answer:
Fault
Explanation:
it only carries a current if there is a fault.
A rifle shoots a 4.20 g bullet out of its barrel. The bullet has a muzzle velocity of just as it leaves the barrel. Assuming a constant horizontal acceleration over a distance of 45.0 cm starting from rest, with no friction between the bullet and the barrel, (a) what force does the rifle exert on the bullet while it is in the barrel? (b) Draw a free-body diagram of the bullet (i) while it is in the barrel and (ii) just after it has left the barrel. (c) How many g’s of acceleration does the rifle give this bullet? (d) For how long a time is the bullet in the barrel?
Answers
(a)The rifle exerts a force of 628 N on the bullet while it is in the barrel.
(b) (i) There is no friction between the bullet and the barrel, there is no force of friction to consider.
(ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.
(c) The rifle gives the bullet an acceleration of about 24885 g's.
(d) the bullet is in the barrel for 0.030 s.
What is Force?
Force is a physical quantity that describes the interaction between two objects, causing a change in motion. Specifically, it is an influence that can cause an object to accelerate, change direction, or deform. Force is defined as the product of mass and acceleration, according to Newton's second law of motion. Mathematically, force can be represented by the equation:
F = m*a
(a) To determine the force that the rifle exerts on the bullet while it is in the barrel, we need to use the equation for the work done by a constant force:
W = Fd
where W is the work done, F is the force, and d is the distance over which the force is applied. We can rearrange this equation to solve for the force:
F = W/d
We know that the work done on the bullet is equal to the change in kinetic energy:
W = ΔK = (1/2)mv^2
where m is the mass of the bullet and v is its velocity. Plugging in the given values, we get:
W = (1/2)(0.00420 kg)(370 m/s)^2 = 283 J
The distance over which the force is applied is given as 45.0 cm = 0.45 m. So the force exerted by the rifle on the bullet is:
F = W/d = 283 J / 0.45 m = 628 N
Therefore, the rifle exerts a force of 628 N on the bullet while it is in the barrel.
(b) (i) The free-body diagram of the bullet while it is in the barrel would show two forces acting on it: the force of the rifle pushing it forward, and the force of gravity pulling it downward. Since there is no friction between the bullet and the barrel, there is no force of friction to consider.
(ii) Just after it has left the barrel, the only force acting on the bullet would be the force of gravity pulling it downward.
(c) We can calculate the acceleration of the bullet using the formula:
a = Δv/t
where Δv is the change in velocity and t is the time for which the acceleration occurs. We know that the initial velocity of the bullet is 370 m/s and that it starts from rest, so the change in velocity is:
Δv = 370 m/s
The distance over which the acceleration occurs is given as 45.0 cm = 0.45 m. Using the formula for distance traveled with constant acceleration, we can find the time for which the acceleration occurs:
d = (1/2)at^2
0.45 m = (1/2)a(t^2)
Solving for t, we get:
t = sqrt(0.9/a)
Plugging this into the equation for acceleration, we get:
a = Δv/t = 370 m/s / sqrt(0.9/a)
Solving for a, we get:
a = 2.44 x 10^5 m/s^2
To express this acceleration in units of g's, we can divide by the acceleration due to gravity:
a/g = 2.44 x 10^5 m/s^2 / 9.81 m/s^2 = 24885
Therefore, the rifle gives the bullet an acceleration of about 24885 g's.
(d) Using the same formula as before for distance traveled with constant acceleration, we can solve for the time the bullet is in the barrel:
d = (1/2)at^2
0.45 m = (1/2)(2.44 x 10^5 m/s^2)t^2
Solving for t, we get:
t = \(\sqrt{(0.00092)}\)s = 0.030 s
Therefore, the bullet is in the barrel for 0.030 s.
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The bullet is in the barrel for 7.05 x 10^-5 seconds. Acceleration can also be caused by changes in direction, such as when an object moves in a circular path.
What is Acceleration?
Acceleration is a fundamental concept in physics and plays an important role in understanding the motion of objects. It is used to describe the behavior of objects in a wide range of applications, from simple everyday situations such as cars accelerating and braking, to more complex phenomena such as the acceleration of particles in particle accelerators or the acceleration of celestial bodies in space.
(a) To find the force exerted by the rifle on the bullet, we can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity (muzzle velocity), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance traveled (45.0 cm = 0.45 m).
Rearranging this equation to solve for acceleration:
a = (v^2 - u^2) / 2s
Plugging in the given values:
a = (1200 m/s)^2 / (2 x 0.45 m) = 3.20 x 10^6 m/s^2
The force exerted by the rifle on the bullet can be found using Newton's second law:
F = ma
where F is the force, m is the mass of the bullet (4.20 g = 0.00420 kg), and a is the acceleration we just calculated:
F = 0.00420 kg x 3.20 x 10^6 m/s^2 = 13,440 N
Therefore, the rifle exerts a force of 13,440 N on the bullet while it is in the barrel.
(b)
(i) Free-body diagram of the bullet while it is in the barrel:
The only force acting on the bullet while it is in the barrel is the force exerted by the rifle, which is directed to the right.
|
|
-->| F
|
|
Once the bullet has left the barrel, it is subject to air resistance, which we will assume acts in the opposite direction to the velocity of the bullet. The force of gravity on the bullet is negligible for this problem.
|
|
<--| F_air
|
|
(c) The acceleration given to the bullet can be expressed in terms of g's by dividing by the acceleration due to gravity, g:
a_g = a / g = (3.20 x 10^6 m/s^2) / 9.81 m/s^2 = 326,000 g's
Therefore, the rifle gives the bullet an acceleration of 326,000 g's.
(d) The time the bullet is in the barrel can be found using the kinematic equation:
s = ut + (1/2)at^2
where s is the distance traveled (0.45 m), u is the initial velocity (0 m/s), a is the acceleration we calculated earlier (3.20 x 10^6 m/s^2), and t is the time the bullet is in the barrel (which we want to find).
Rearranging and solving for t:
t = sqrt(2s/a) = sqrt(2 x 0.45 m / 3.20 x 10^6 m/s^2) = 7.05 x 10^-5 s
Therefore, the bullet is in the barrel for 7.05 x 10^-5 seconds.
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Which one of the following is the longest length?
(a) 100 m
(b) 104 µm
(c) 107 nm
(d) 102 mm
Answers
Okay, let's convert all the lengths to the same unit to compare:
(a) 100 m = 100 meters
(b) 104 μm = 104 micrometers = 104 × 10^-6 meters = 0.000104 meters
(c) 107 nm = 107 nanometers = 107 × 10^-9 meters = 0.000000000997 meters
(d) 102 mm = 102 millimeters = 102 × 10^-3 meters = 0.0102 meters
The longest length is:
(a) 100 m = 100 meters
The answer is option (a).
Answer: 100 m
Explanation:
1 μm = \(10^{-6}\) m = 0,000001 m
1 nm = \(10^{-9}\) m = 0,000000001 m
1 mm = \(10^{-3}\) m = 0,001 m
∴ 100 m es la mayor longitud
Particles q₁ +8.0 μC, q2 +3.5 μC, and
93-2.5 μC are in a line. Particles q₁ and q2 are
separated by 0.10 m and particles q2 and q3 are
separated by 0.15 m. What is the net force on
particle q₂?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
+8.0μ.C
+91
0.10 m
+3.5 C
+92
0.15 m
-2.5μ C
93
Answers
The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.
To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.
The force between two charged particles can be calculated using Coulomb's law:
F = k * |q₁ * q₂| / r²
Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.
First, let's calculate the force exerted on q₂ by q₁:
F₁₂ = k * |q₁ * q₂| / r₁₂²
F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²
F₁₂ ≈ 252000 N
The force is positive because q₁ and q₂ have opposite charges.
Next, let's calculate the force exerted on q₂ by q₃:
F₂₃ = k * |q₂ * q₃| / r₂₃²
F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²
F₂₃ ≈ -63000 N
The force is negative because q₂ and q₃ have the same charge.
Finally, we can find the net force on q₂ by summing the individual forces:
Net force = F₁₂ + F₂₃
Net force = 252000 N + (-63000 N)
Net force ≈ 189000 N
The net force on particle q₂ is approximately 189000 N.
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A flat sheet of paper of area 0.450 m2 is oriented so that the normal to the sheet is at an angle of 600 to a uniform electric field of magnitude 18 N C-1. What is the magnitude of the electric flux through the sheet? A. 3.22 N m2 C-1 B. 21.42 N m2 C-1 C. 5.04 N m2 C-1 D. 11.72 N m2 C-1 E. 4.05 N m2 C
Answers
The magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹ (Option E).
The electric flux through a surface is given by the product of the electric field strength and the area of the surface projected perpendicular to the electric field.
In this case, the electric field strength is 18 N C⁻¹, and the area of the sheet projected perpendicular to the electric field is 0.450 m²
(since the normal to the sheet makes an angle of 60° with the electric field). Multiplying these values gives the electric flux:
Electric flux = Electric field strength × Area
Electric flux = 18 N C⁻¹ × 0.450 m²
Electric flux = 8.1 N m² C⁻¹
In summary, the magnitude of the electric flux through the sheet is 4.05 N m² C⁻¹. This value is obtained by multiplying the given electric field strength by the projected area of the sheet perpendicular to the electric field.
The angle of 60° is taken into account to determine the effective area for calculating the flux.(Option E).
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A 80 kg
bicyclist is coasting down a long hill that has a 5.0 ∘
slope. His cross-section area is 0.35 m2
, his drag coefficient is 0.85, his bicycle's coefficient of rolling friction is 0.02, and the air temperature is 20∘C
.
Answers
the Speed reached by the bicyclist is given by, v = 26 m/s
What are the types of friction?There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. Friction and normal force are directly proportional to the contacting surfaces, and it doesn't depend on the hardness of the contacting surface.
Given that,
Gravitational acceleration = 9.81 m/s²
Mass of bicyclist = 80 kg
Slope(θ) = 5°
Cross-sectional area of bicyclist = 0.35 m²
μ = coefficient of rolling friction = 0.02
t = air temperature = 20 °C
Let ρ be the density of air at 20oC = 1.2041 kg/m3
ρ = 1.2041 kg/m3
We have the balancing equation:
F₁ = F₂ + F₃
where,
F₁ = component of the weight of the bicyclist down the plane,
F₂ = frictional force up the plane, and
F₃ = air drag force up the plane.
Thus, F₃ = F₁ - F₂
sinθ = sin(5°) = 0.087
cosθ = cos(5°) = 0.996
Now we will calculate the forces,
F₁ = m g sinθ = 80 kg x 9.81 m/s2 x 0.087 = 68.227 N
F₁ = 68.227 N
F₂ = μ m g cosθ = 0.02 x 80 kg x 9.81 m/s2 x 0.996 = 15.63 N
F₂ = 15.63 N
Put the values of F1 and F2 in eqn (1) and we get:
F₃ = F1 - F2
F₃ = 68.227 - 15.63
F₃ = 52.597 N
Let v be the speed reached by a bicyclist,
From the general formula of velocity:
V = ((2 xF₃ / (ρ x CD x A))1/2
v = ((2 x F₃ / (ρ x CD x A))1/2
v = ((2 x 120.78 N / (1.2041 kg/m3 x 0.85 x 0.35))1/2 = 25.97 m/s
v = 26 m/s
Therefore, the Speed reached by the bicyclist is given by, v = 26 m/s
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How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another?
Answers
Answer:
94,800 m
Explanation:
F = kq1 q2/r^2
1 = 9 x 10^9 x 1 / r^2
A__is the best example or representation of a concept.
Answers
A prototype is the best example or representation of a concept, which is used as a design to shape a device.
What is a prototype?A prototype can be defined as a given design in its experimental stage which is very useful to shape the framework or idea to carry out.
Therefore, with this data, we can see that a prototype can be very useful to shape an idea or concept about a given device in particular.
Complete question:
Fill in the blank. A _____ is the best example or representation of a concept.
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What has holes but can still hold water?
Answers
anaerobic transmission is when you touch a contaminated surface true or false
Answers
Anaerobic transmission is when you touch a contaminated surface is a false statement.
What is anaerobic respiration?Anaerobic organisms are the living things that can survive and grow where there is no oxygen in the surrounding environment so we can conclude that Anaerobic transmission is when you touch a contaminated surface is a false statement.
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the distance covered by a car at a time t is given by x = 20t + 6t^2 . calculate the instantaneous velocity and acceleration when t = 1s
Answers
The velocity is 22 m/s while the acceleration is 12 m/s^2.
What is the instantaneous velocity?We know that the velocity can be obtained from the equation that have been given for the distance that is covered. We know that we have been given the distance that is covered as; x = 20t + 6t^2
Then we know that;
Velocity = dx/dt = 20 + 12t
Velocity = 20 + 12(1) = 22 m/s
The acceleration can be obtained from;
Acceleration = d^2x/dt^2 = 12 m/s^2
Thus we have used the equation to obtain the velocity and the acceleration.
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prepare a report on why a vehicle needs to be maintained/serviced after a certain period of time. How is servicing different in a petrol/diesel and electric vehicle?
Answers
Vehicles need to be serviced for several reasons such as preventing costly repairs and improving fuel economy.
Why should cars be maintained and / or serviced ?First, regular maintenance can help to prevent costly repairs down the road. Second, maintenance can help to improve fuel economy and emissions. Third, maintenance can help to keep your vehicle safe and reliable.
The servicing requirements for petrol/diesel and electric vehicles differ in a number of ways. Petrol/diesel vehicles require oil changes more frequently than electric vehicles. This is because petrol/diesel engines use oil to lubricate the moving parts, while electric motors do not. Petrol/diesel vehicles also require tune-ups more frequently than electric vehicles.
This is because petrol/diesel engines have more moving parts that need to be synchronized, while electric motors have fewer moving parts.
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the order to stop given at time t=0s state the speed of the car at t=0s
Answers
A sperm whale can accelerate at about 0.100 m/s2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 1.10 m/s and accelerates to a speed of 2.44 m/s? Assume the whale travels in a straight line.
Answers
The distance traveled by the sperm whale traveling at an initial speed of 1.10m/s to 2.44m/s is 23.7m.
How to calculate distance?The distance traveled by the sperm whale can be estimated using one of the equations of motion as follows:
v² = u² + 2as
Where;
v = final velocityu = initial velocitya = accelerations = distanceAccording to this question, a sperm whale can accelerate at about 0.100 m/s² when swimming on the surface of the ocean. The distance traveled by the whale can be calculated as follows:
2.44² = 1.10² + 2(0.10)(s)
5.95 = 1.21 + 0.2s
4.74 = 0.2s
s = 23.7m
Therefore, 23.7m is the distance of the sperm whale.
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c. Boat travels north then west
A boat travels 76.0 km due north in 8.0 hours then 56.0 km due west in 5.0 hours.
Determine the direction (as a bearing) of the average velocity (to 1 decimal places) of the boat in the 8 + 5 hour period.
PLEASE HELP!!
Answers
Answer:
Explanation:
θ = arctan(56.0/76.0) = 36.4° West of North
average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s
How many pets do you have???!!!??
Answers
Answer:
Right now I have three.
Explanation: Thanks for the points luv ^-^.
Answer:
I have two
Explanation:
What type of circuit is shown
Answers
=When three resistors are connected in parallel, they form a parallel circuit. In a parallel circuit, each resistor is connected across the same two points, with the current split between the resistors.
In this configuration, the voltage across each resistor is the same, but the current through each resistor can be different. The total resistance of the circuit is calculated using the equation:
1/R_total = 1/R1 + 1/R2 + 1/R3
where R1, R2, and R3 are the resistance values of the individual resistors.
The total current in the circuit is equal to the sum of the currents through each resistor:
I_total = I1 + I2 + I3
where I1, I2, and I3 are the currents through each resistor.
The total power dissipated in the circuit can be calculated using the equation:
P_total = V² / R_total
where V is the voltage across the resistors.
In summary, when three resistors are connected in parallel, they form a parallel circuit, with each resistor connected across the same two points, and the current split between them. The total resistance, current, and power dissipated in the circuit can be calculated using the equations provided.
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The container contains
60 °C 500 ml of water.
Add 0 °C of ice 300 g.
Calculate the final
temperature of the
system The heat capacity
of the vessel is 45 kJ /
°C.
Answers
A) calculate the extension of wire needed to produce this tensionBCalculate the work done in producing this extension
Answers
Given,
The length of the string, L=0.85 m
The diameter of the string, d=6×10⁻⁴ m
The radius of the string, r=3×10⁻⁴ m
The Young's Modulus of the string, Y=190 GPa
The tension, F=66 N
a)
The Young's Modulus is given by,
\(Y=\frac{FL}{Al}\)Where A is the area of cross-section of the string and l is the extension produced in the string.
On rearranging the above equation,
\(l=\frac{FL}{YA}\)On substituting the known values,
\(\begin{gathered} l=\frac{66\times0.85}{190\times10^9\times\pi(3\times10^{-4})^2} \\ =1.04\times10^{-3}\text{ m} \end{gathered}\)Thus the extension of the wire that is needed to be produced is 1.04×10⁻³ m
b)
The work done in producing the extension is given by,
\(W=\frac{1}{2}Fl\)On substituting the known values,
\(\begin{gathered} W=\frac{1}{2}\times66\times1.04\times10^{-3} \\ =0.034\text{ J} \end{gathered}\)Thus the work done in producing the extension is 0.034 J
Which is usually associated with a faster reaction rate
Answers
Answer:
If there are more molecules present, or there's a bigger surface area on which the reaction happens, there will be more successful collisions and the reaction will go faster. Also, if the temperature is higher, more molecules will have enough energy to react, and the reaction will be faster.
Light travels about 180 million kilometers in 10 minutes. How far does it travel in 1 minute? How far does it travel in 1 second? Show your reasoning
I need help
Answers
Explanation:
180 million km = 10 min
? = 1 min
180 million x 10 = 1,800,000,000 km
180 million km = 600s
? = l s
108,000,000,000km
Download the attached Word Document and follow the instructions to complete the lab. Please be sure to to answer all questions.
Copy and paste your answer in the box below or attach a Word Document (.doc or .docx) or PDF (.pdf).
Answers
Topic: The Effect of Temperature on Enzyme Activity in Plants
Hypothesis: I predict that as the temperature increases, the rate of enzyme activity in plants will also increase. As the temperature decreases, the rate of enzyme activity in plants will decrease.
Temperature is the measure of the average kinetic energy of the particles in a substance. It is measured in degrees Celsius (°C), Fahrenheit (°F), or Kelvin (K). Temperature is determined by the amount of heat present in a substance, which is a measure of the average kinetic energy of its particles. Heat is the energy transferred from a hotter object to a cooler one. When two objects of different temperatures come into contact, heat will flow from the hotter object to the cooler one until both objects reach the same temperature. Temperature can be affected by many factors, such as the weather, the amount of sunlight, and the activity of the particles in a substance.
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A 2.15-nC charged particle located at the origin is separated by a distance of 0.0690 m from a 3.78-nC charged particle located farther along the positive x axis. If the 2.15-nC particle is kept fixed at the origin, where along the positive x axis should the 3.78-nC particle be located so that the magnitude of the electrostatic force it experiences is twice as great as the magnitude of the electrostatic force it experiences at 0.0690 m?
Answers
The new distance of the 3.78 nC particle when the electrostatic force is doubled is 0.049 m.
What is the electrostatic force between the two particles?
The magnitude of the electrostatic force experienced by the 3.78 n C particle when it is 0.069 m from the 2.15 nC particle is calculated as follows;
F = kq₁q₂ / r²
where;
k is the Coulomb's constantq₁ is the magnitude of particle 1q₂ is the magnitude of particle 2r is the distance between the particlesF = ( 9 x 10⁹ x 2.15 x 10⁻⁹ x 3.78 x 10⁻⁹ ) / ( 0.069² )
F = 1.536 x 10⁻⁵ N
When this electrostatic force is doubled, the new distance of 3.78 nC particle from the 2.15 nC is calculated as;
r² = kq₁q₂ / 2F
r = √ ( kq₁q₂ / 2F )
r = √ [ ( 9 x 10⁹ x 2.15 x 10⁻⁹ x 3.78 x 10⁻⁹ ) / ( 2 x 1.536 x 10⁻⁵ ) ]
r = 0.049 m
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If you converted 0.000013 to scientific notation, what would the prefix be to the correct number of significant digits?
Answers
Answer:
1.3
Explanation:
it will taken to as in from of standard form
Answer:
1.3 * 10^-5
Explanation:
We are learning about scientific notation.
When a number becomes a decimal followed before with zeroes, we know that the value of that number is decreasing. So instead of usually doing a positive exponent, we will do a negative exponent indicating we are going back.
So let's not only count the amount of zeroes followed before 13, but the decimal.
0.000013
The original number "1.3" went back 5 spaces, therefore making our exponent 5.
1.3 * 10^-5
how is heating up your hands by a fire not an example of conduction?
Answers
Let's explain how heating up your hands by a fire is not an example of conduction.
Conduction can be said to be the transfer of heat which involves the movement of particles which are in contact with each other.
In conduction there is no transfer material.
So, when we heat up our hands with fire, the air acts as a thermal insulator. Since the air acts as a thermal insulator, this means there is now a transfer material.
As the fire burns, it heats the air which in turn expands and rises outwards causing your hand to feel the warm air.
The type of heat transfer that occurs when you heat you hands by a fire is convection. The heat flows from a higher temperature to a lower temperature.
ANSWER:
Therefore, heating up your hands by a fire is not an example of conduction because the air already acts as a thermal insulator
A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
Answers
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe upward force exerted on the board by the support is calculated as follows;
F(up) = 52.8 N + 206.0 N + 272.0 N
F(up) = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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Please help with this question
Answers
Answer:
I believe the answer is 5718.75. Respond if it is wrong please.
Explanation:
I used a calculator.
A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest. What is the common velocty of the two-car train after the collisions?
Answers
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
\(m_1u_1+m_2u_2=(m_1+m_2)V\)
V is the common velocity after the collisions
\(V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s\)
So, the two car train will move with a common velocity of 6.79 m/s.
Two capacitors 6 micro farad and 8 micro farad are connected in parallel. The combination is then connected in series with a 12 V battery and 14 micro farad capacitor. What is the charge on the 6 micro farad capacitor?
Answers
The charge on the 6 μF capacitor is 10.29 μC.
What is the charge of capacitor?
The charge of a capacitor is the amount of electric charge stored on its plates due to the potential difference (voltage) between them. The charge is directly proportional to the capacitance and the voltage across the capacitor, according to the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
First, we can calculate the equivalent capacitance of the two capacitors connected in parallel:
\(C_{parallel}\)= C1 + C2 = 6 μF + 8 μF = 14 μF
Next, we can calculate the total equivalent capacitance of the circuit, which is the capacitance of the parallel combination in series with the 14 μF capacitor:
1/\(C_{total}\) = 1/ \(C_{parallel}\) + 1/C3
1/\(C_{total}\) = 1/14 μF + 1/14 μF
1/\(C_{total}\) = 1/7 μF
\(C_{total}\) = 7 μF
Using the formula Q = CV, we can calculate the charge on the 6 μF capacitor:
Q = C × V = 6 μF × 12 V / 7 μF = 10.29 μC
Therefore, the charge on the 6 μF capacitor is 10.29 μC.
What is capacitance?
Capacitance is the ability of a capacitor to store electric charge when a voltage is applied across its terminals. It is defined as the ratio of the electric charge stored in the capacitor to the voltage applied to it, and its unit is the farad (F). A capacitor with a high capacitance value can store more charge for a given voltage than a capacitor with a lower capacitance value.
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A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car moves a distance of 2.0 meters. How much additional distance will the car move during the second second of its motion?
Answers
Since the car is accelerating at a constant rate, the distance it travels during each second of its motion will be directly proportional to the time it has been accelerating.
In the first second, the car moved a distance of 2 meters, and in the second second, it will move twice the distance of the first second, so the car will move additional distance of 2*2 = 4 meters during the second second of its motion.
The distance traveled during the second second of its motion is 1/2 * 2 = 1 meters.
A car that accelerates at a constant rate will move a distance equal to the initial velocity multiplied by time plus 1/2 the acceleration multiplied by the square of time. Since the car starts from rest, the initial velocity is zero.
Therefore, the distance traveled during the second second is 1/2 * acceleration \(* (time)^2 = 1/2 * a * t^2 = 1/2 * a * 1^2 = 1/2 * a\) Since the car moved 2.0 meters in the first second, it means the acceleration is\(2m/s^2\), and the distance traveled during the second second is 1/2 * 2 = 1 meters.
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