Answer:
The new separation is [tex]\bf{(d/\sqrt{3})}[/tex].
Explanation:
The expression of the force between two spheres is given by
[tex]F = k\dfrac{q_{A}q_{B}}{d^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where, [tex]q_{A}[/tex] is the charge on sphere A, [tex]q_{B}[/tex] is the charge on sphere B, [tex]k[/tex] is constant and [tex]d[/tex] i the separation between two spheres.
The new value of charge on sphere B is [tex]q_{B}^{n} = \dfrac{q_{A}}{3}[/tex]. Consider the new separation between the spheres be [tex]d'[/tex]. Under the new configuration the force between the spheres is given by
[tex]F = k \dfrac{q_{A}(q_{A}/3)}{d'^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Equating equation (1) and equation (2), we have
[tex]~~~~&& \dfrac{1}{d'^{2}} = \dfrac{1}{3d^{2}}\\&or,& d' = \dfrac{d}{\sqrt{3}}[/tex]
So, the new separation is [tex](d/\sqrt{3})[/tex].
If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be pa
Answer:
The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is [tex]\frac{135.9}{A}N[/tex]
Explanation:
Here we have
Initial temperature of air T₁ = 20 °C = 293.15 K
Final temperature of air T₁ = 120 °C = 393.15 K
Initial pressure P₁ = 1 atm = 101325 Pa
Final pressure P₂ = Required
Area = A
Therefore we have for the pressure cooker, the volume is constant that is does not change
By Chales law
P₁/T₁ = P₂/T₂
P₂ = T₂×P₁/T₁ = 393.15 K× (101325 Pa/293.15 K) = 135,889.22 Pa
∴ P₂ = 135.88922 KPa = 135.9 kPa
Where Force = [tex]\frac{Pressure}{Area}[/tex] we have
Force = [tex]F_{120}=\frac{135.9}{A}N[/tex].
how does an antenna produce radio waves?
Answer:
As the electrons (tiny particles inside atoms) in the electric current wiggle back and forth along the antenna, they create invisibleelectromagnetic radiation in the form ofradio waves. ... 1) Electricity flowing into the transmitter antenna makes electrons vibrate up and down it, producing radio waves.
Answer:
As the electrons (tiny particles inside atoms) in the electric current wiggle back and forth along the antenna, they create invisible electromagnetic radiation in the form of radio waves.
Explanation:
In the space below, explain why you agree or disagree with the first statement: Each person in a family has the same traits. There are no differences in traits between parents and offspring or among siblings.
Answer:
I disagree.
Explanation:
Yes, traits may be similar, but it all depends on the dominant and recessive alleles that are passed on. No one person can look alike. Even with twins, a widow's peak or close lobes can be different.
I hope this was the brainliest answer! Thank you for letting me help you.
Two technicians are discussing a problem where the brake pedal travels too far before the vehicle starts to slow. Technician A says that the brakes may be out of adjustment. Technician B says that one circuit from the master cylinder may be leaking or defective. Which technician is correct?
Answer:
Technician A
Explanation:
If Technician B was correct, and the master cylinder is defective - then no braking action would occur.
This is not true however, as some breaking action eventually occurs, meaning it must be out of adjustment.
Both Technician A, suggesting the brakes are out of adjustment, and Technician B, suggesting a leak or defect in the master cylinder, could be correct in the scenario of extended brake pedal travel before the vehicle slows.
Explanation:Both Technician A and Technician B could be correct in diagnosing a problem where the brake pedal travels too far before the vehicle starts to slow down. Technician A suggests that the brakes may be out of adjustment. If the brakes are not properly adjusted, the brake pads or shoes may be too far from the rotor or drum, causing the pedal to travel further before the pads make contact and slow the vehicle.
Technician B considers a hydraulic issue, proposing that one circuit from the master cylinder may be leaking or defective. In a hydraulic brake system, if there is a leak or a defect in one of the cylinders, it could result in a loss of pressure when the brake pedal is applied. This loss of pressure means the braking force is not adequately transmitted to the brake pads, leading to increased pedal travel.
Hydraulic brakes use Pascal's principle, where pressure applied to a confined fluid is transmitted undiminished in all directions. The master cylinder, when the brake pedal is applied, generates pressure that is transferred to the slave cylinders located at each wheel. If the master cylinder is compromised or out of adjustment, the result is insufficient pressure and force at the slave cylinders, hence longer pedal travel before effective braking occurs.
Describe what is happening within the system when it is at equilibrium in terms of concentrations
Answer:
When a system is at equilibrium in terms of concentration what happens is that the rate of change of the concentration of the product and the reactants does not vary or change with time.
Explanation:
What is equilibrium?
A chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.
Equilibrium does not necessarily mean that reactants and products are present in equal amounts. It means that the reaction has reached a point where the concentrations of the reactant and product are unchanging with time, because the forward and backward reactions have the same rate.
Final answer:
A chemical system at equilibrium reflects no net change in reactant and product concentrations. Upon a concentration change, Le Chatelier's principle dictates that the system will adjust to partially counteract the change and set up a new equilibrium.
Explanation:
When a system is at equilibrium, it means no net change in the concentrations of reactants and products takes place. This is because the rate of the forward reaction, in which reactants are converted into products, is equal to the rate of the reverse reaction, in which products transform back into reactants. However, when there is a change in concentration of either reactants or products during equilibrium, the system responds according to Le Chatelier's principle. This principle states that the system will adjust to partially counteract the change and reestablish a new equilibrium state.
For example, if you were to increase the concentration of a reactant, the system tends to counter this by producing more product, essentially shifting the equilibrium to the right. Conversely, if you decrease the concentration of a product, the system will proceed to produce more of that product from the reactants, again shifting the reaction to the right, until a new equilibrium is established.
Which of the following treatments would enhance the level of the Pfr form of phytochrome?A) exposure to far-red lightB) exposure to red lightC) long dark periodD) inhibition of protein synthesisE) synthesis of phosphorylating enzymes
Answer:
B) exposure to red light
Explanation:
Plants use a phytochrome system to sense the level, intensity, duration, and color of environmental light do as to adjust their physiology.
The phytochromes are a family of chromoproteins with a linear tetrapyrrole chromophore, similar to the chlorophyll. Phytochromes have two photo-interconvertible forms: Pr and Pfr. Pr absorbs red light (~667 nm) and is immediately converted to Pfr. Pfr absorbs far-red light (~730 nm) and is quickly converted back to Pr. Absorption of red or far-red light causes a massive change to the shape of the chromophore, altering the conformation and activity of the phytochrome protein to which it is bound. Together, the two forms represent the phytochrome system.
The magnetic field within a long, straight solenoid with a circular cross section and radius r is increasing at a rate of dbdt. part a what is the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? express your answer in terms of the variables r, b, r1, and appropriate constants.
Answer:
[tex]\frac{d\Phi_B}{dt}=\frac{d(\pi r_1^2B)}{dt}=\pi r_1^2\frac{dB}{dt}[/tex]
Explanation:
To calculate the rate of change of the flux we have to take into account that the magnetic flux is given by
[tex]\Phi_B=\vec{B}\cdot \vec{A}[/tex]
in this case the direction of B is perpendicular to the direction of A. Hence
[tex]\Phi_B=BA[/tex]
and A is the area of a circle:
[tex]A=\pi r^2[/tex]
in this case we are interested in the flux of a area of a lower radius r1. Hence
[tex]A=\pi r_1^2[/tex]
Finally, the change in the magnetic flux will be
[tex]\frac{d\Phi_B}{dt}=\frac{d(\pi r_1^2B)}{dt}=\pi r_1^2\frac{dB}{dt}[/tex]
hope this helps!!
Four long, parallel conductors carry equal currents of I = 5.00 A. The direction of the current is into the page at points A and B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.200 m.
Answer:
Explanation:
The magnetic field due to a straight wire carrying current is given by
B = μo• I / 2πr
Where,
μo = permeability of free space
μo = 4π×10^-7 Tm/A
I is the current in wire
I = 5A
r Is the distance of point from the wire.
The distance between the parallel conductors and the point is r/2
R = 0.283/2 = 0.1415m
Check attachment on how I used Pythagoras theorem to find the diagonal of the square..
Hence, the magnetic field at point P is
B = μo•I / 2πR
B = 4π × 10^-7 × 5 / 2π × 0.1415
B = 7.07 × 10^-6 T.
50 kg grandma is roller skating down the sidewalk at 2 m/s. Suddenly, someone throws her a 10 kg sack of oranges. How fast does she travel afterwards? (Inelastic)
Answer:
v2 = 1.67m/s
she will travel at 1.67m/s afterwards
Explanation:
Given;
Mass of grandma m1 = 50kg
Mass of sack m2 = 10kg
Initial speed of grandma v1 = 2m/s
Final velocity of both = v2
The momentum equation of the initial and final momentum can be written as;
m1v1 = m1v2 + m2v2 = (m1+m2)v2
(Since the collision is inelastic they both move at the same velocity v2 after the collision)
making v2 the subject of formula;
v2 = (m1v1)/(m1+m2)
Substituting the values;
v2 = (50×2)/(50+10)
v2 = 1.67m/s
she will travel at 1.67m/s afterwards
The time between a lightning flash and the following thunderclap may be used to estimate, in kilometers, how far away a storm is. How far away is a storm if 6 seconds elapse between the lightning and the thunderclap?
Given Information:
Elapsed time = t = 6 seconds
Required Information:
Distance = d = ?
Answer:
Distance = d = 2.058 km
Explanation:
We know that the speed of sound in the air is given by
v = 343 m/s
The relation between distance, speed and time is given by
distance = speed*time
substituting the given values yields,
distance = 343*6
distance = 2058 m
There are 1000 meters in 1 km so
d = 2058/1000
d = 2.058 km
Therefore, the storm is about 2.058 km away when elapse time between the lightning and the thunderclap is 6 seconds.
In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is
Fext=dp/dt=m dv/dt+v dm/dt
Suppose the mass of the raindrop depends on the distance x that it has fallen. Then m = kx, where k is a constant, and dm/dt=kv
dm/dt=kv This gives, since Fext=mg
Fext=mg,
mg=m dv/dt+v(kv)
Or, dividing by k,
xg=x dv/dt+v2
This is a differential equation that has a solution of the form v = at, where a is the acceleration and is constant. Take the initial velocity of the raindrop to be zero.
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
The acceleration of the raindrop is equal to the acceleration due to gravity. The distance the raindrop has fallen in 3.00 seconds is 44.1 meters. The mass of the raindrop at 3.00 seconds is 88.2 grams.
Explanation:To find the acceleration, we can use the given proposed solution for velocity, v = at. Taking the derivative of this equation with respect to time gives us dv/dt = a. Substituting this into the differential equation, we have xg = x(a) + v^2. Since the initial velocity is zero, v = 0 and the equation simplifies to xg = xa. Dividing by x, we get g = a. Therefore, the acceleration is equal to the acceleration due to gravity, g.
To find the distance the raindrop has fallen in t = 3.00 s, we can use the equation x = (1/2)at^2. Since we know the acceleration is g, we plug in the values into the equation: x = (1/2)(g)(3.00 s)^2 = (1/2)(9.8 m/s^2)(9.00 s^2) = 44.1 meters.
To find the mass of the raindrop at t = 3.00 s, we can use the given equation m = kx. Plugging in the values, m = (2.00 g/m)(44.1 m) = 88.2 grams.
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A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.18 s?
Answer:
4.7 N
Explanation:
130 g = 0.13 kg
The momentum of the snowball when it's thrown at the wall is
[tex]p = mv = 0.13*6.5 = 0.845 kgm/s[/tex]
Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s
[tex]p = F\Delta t[/tex]
[tex]F*0.18 = 0.845[/tex]
[tex]F = 0.845 / 0.18 = 4.7 N[/tex]
A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of atoms are there in the sheet?
Final answer:
To calculate the number of gold atom layers in a 0.125 micrometer thick gold leaf, convert the thickness to picometers and divide by the diameter of a gold atom, which is twice its radius of 174pm, resulting in approximately 359 layers.
Explanation:
The question "A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of atoms are there in the sheet?" is asking us to calculate the number of gold atom layers in a given thickness of a gold leaf.
First, we need to convert the given thickness of the gold leaf from micrometers to picometers to match the atomic radius unit. There are 1,000,000 picometers in a micrometer, so 0.125 micrometers is equal to 125,000 picometers. Since the diameter of a gold atom is twice the radius, we have a diameter of 174pm x 2 = 348pm. Now, to find out how many atoms can fit in the thickness, we divide the total thickness by the diameter of one atom:
125,000 pm / 348 pm ~= 359 layers of atoms
Therefore, there are approximately 359 layers of gold atoms in the sheet of gold leaf.
There are approximately 20 layers of atoms in the sheet of gold leaf.
To determine the number of layers of atoms in the sheet of gold leaf, we need to compare the thickness of the sheet with the diameter of a single gold atom.
First, we convert the thickness of the gold leaf from micrometers to picometers (pm) to match the units of the gold atom's diameter.
Since 1 micrometer is equal to 1000 picometers, the thickness of the gold leaf in picometers is:
[tex]\[ 0.125 \times 1000 = 125 \text{ pm} \][/tex]
Next, we need to consider the diameter of a gold atom, which is twice the radius.
Given that the radius is 174 pm, the diameter is:
Using the FCC packing efficiency of 74%, we get:
[tex]\[ \frac{1}{0.74} \approx 1.35 \][/tex]
Again, rounding up to the nearest whole number, we get 2 layers.
Therefore, considering the packing efficiency of gold atoms in the sheet, there are approximately 20 layers of atoms in the sheet of gold leaf.
Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Explanation:
Let us assume that the mass of a pitched ball is 0.145 kg.
Initial velocity of the pitched ball, u = 47.5 m/s
Final speed of the ball, v = -51.5 m/s (in opposite direction)
We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :
[tex]\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s[/tex]
So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.
Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :
[tex]J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s[/tex]
Hence, this is the required solution.
The magnitude of the change in momentum of the ball is 99 m/s. The magnitude of the impulse applied to the ball by the bat is also 99 m/s.
Explanation:To find the magnitude of the change in momentum of the ball, we can use the equation:
Magnitude of change in momentum = magnitude of final momentum - magnitude of initial momentum
Given that the initial velocity of the ball is 47.5 m/s and the final velocity of the batted ball is 51.5 m/s in the opposite direction, the magnitude of the change in momentum is:
Magnitude of change in momentum = 51.5 m/s - (-47.5 m/s) = 99 m/s
The magnitude of the impulse applied to the ball by the bat is equal to the magnitude of the change in momentum. Therefore, the magnitude of the impulse applied to the ball is 99 m/s.
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What is the energy difference between the h2 molecule and the separated atoms?
Answer:My answer its a little large so
Explanation:
i write in a paper for you i believe you can be able to read it good
The bond energy is the difference in energy between a molecule and its separated atoms. For the H2 molecule, the energy difference is 7.24 × 10-19 J at a bond distance of 74 pm.
Explanation:The energy difference between the H2 molecule and the separated hydrogen atoms is the bond energy, which is the difference between the energy minimum at the bond distance and the energy of the separated atoms. For the H2 molecule, at a bond distance of 74 pm, the system is lower in energy by 7.24 × 10-19 J compared to the separated atoms. This small energy difference is important for the stability of the H2 molecule.
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An electric circuit consists of a variable resistor connected to a source of constant potential difference. If the resistance of the resistor is doubled, then
A ball thrown upward near the surface of the Earth with a velocity of 50 m/s will come to rest about 5 seconds later. If the ball were thrown up with the same velocity on Planet X, after 5 seconds it would be still moving upwards at nearly 31 m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the Earth? Using the MathType functions, be sure to show your work
Answer:
19/49
Explanation:
Using v = u + at where v = velocity of ball after 5 s on planet X = 31 m/s, u = initial velocity of ball on planet X = 50 m/s , a = acceleration due to gravity on planet X and t = 5 s
So, 31 = 50 - a × 5 = 50 - 5a
31 - 50 = 5a
-19 = 5a
a = -19/5 = -3.8 m/s²
So, the magnitude of a = 3.8 m/s²
So a/g = 3.8/9.8 = 19/49
The fraction of the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth is 0.39.
Given the following data:
Time = 5 secondsInitial velocity = 50 m/sFinal velocity = 30 m/sWe know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]
To determine what fraction is the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth:
First of all, we would calculate the acceleration due to gravity (g) on Planet X by using first equation of motion:
Mathematically, the first equation of motion is calculated by using the formula;
[tex]V = U-at[/tex]
Where:
V is the final velocity.U is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]31=50-a(5)\\\\5a =50-31\\\\5a=19\\\\a=\frac{19}{5}[/tex]
Acceleration, a = 3.8 [tex]m/s^2[/tex]
For the ratio:
[tex]Fraction = \frac{a}{g} \\\\Fraction = \frac{3.8}{9.8}[/tex]
Fraction = 0.39
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Which category of materials allows electrons to travel through them freely
The idea that little children are continually coming up with ideas and testing them is called the ______ theory.
atomic
theory
moral foundations
general
Answer:
The idea that little children are continually coming up with ideas and testing them is called the general theory.
A parallel-plate capacitor has plates of area 0.30 m2 and a separation of 2.10 cm. A battery charges the plates to a potential difference of 190 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates.
(a) What is the capacitance before the slab is inserted?
(b) What is the capacitance with the slab in place?
(c) What is the free charge q before and after the slab is inserted?
(d) What is the magnitude of the electric field in the space between the plates and dielectric and in the dielectric itself?
(f) With the slab in place, what is the potential difference across the plates?
(g) How much external work is involved in inserting the slab?
Answer:
a) 1.26e^-10F
b) 1.47e^-10F
c) 2.39e^-8C 2.89e^-8C
d) E=4500.94N/C
e) E'=5254.23N/C
f) 100.68V
g) 1.65e^-10J
Explanation:
To compute the capacitance we can use the formula:
[tex]C=\frac{k\epsilon_o A}{d}[/tex]
where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.
(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:
[tex]k=1\\A=0.30m^2\\d=0.021m\\e_o=8.85*10^{-12}C^2/(Nm^2)\\\\C=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{0.021m}=1.26*10^{-10}F[/tex]
(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:
[tex]C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1})\\\\\\C_1=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{8.5*10^{-3}m}=3.1*10^{-10}F\\\\C_2=\frac{(4.8)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{4*10^{-3}m}=3.186*10^{-9}F\\\\C=1.47*10^{-10}F[/tex]
(c)
The charge between the plates for both cases, with the slab is given by:
Q : without the slab
Q': with the slab
[tex]Q=CV=(1.26*10^{-10}F)(190V)=2.39*10^{-8}C\\\\Q'=C'V=(1.47*10^{-10F})(190V)=2.79*10^{-8}C\\[/tex]
(d) The electric field between the plate is given by:
[tex]E=\frac{Q}{2\epsilon_o A}[/tex]
E: without the slab
E': with the slab
[tex]E=\frac{2.39*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=4500.94N/C\\\\E'=\frac{2.79*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=5254.23N/C\\[/tex]
(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:
[tex]V=2(4500.94)(8.85*10^{-3}m)+(5254.23)(4*10^{-3}m)=100.68V[/tex]
(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels:
[tex]\Delta E=\frac{1}{2}[(1.26*10^{-10}F)(120V)-(1.47*10^{-10})(100.6V)]=1.65*10^{-10}J[/tex]
A(n)______is a device intended for the protection of personnel that functions to deenergize a circuit or portion thereof within an established period of time when a current to ground exceeds the values established for a Class A device.
Answer:
The answer is GFCI
(ground fault circuit interrupter)
Explanation:
A ground fault circuit interrupter (GFCI) can help prevent electrocution.
If a person's body starts to receive a shock, the GFCI senses this and cuts off the power before he/she can get injured. GFCIs are generally installed where electrical circuits may accidentally come into contact with water.
What is a circuit breaker?
It is an automatically operated electrical switch designed to protect an electrical circuit from damage caused by excess current from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteries must be able to store and move in order to accelerate the 770 kg car from rest to 26 m/s, make it climb a 2.15 x 10^2 m high hill while maintaining that speed, and then cause it to travel at a constant 26 m/s by exerting a 5.3 x 10^2 N force for an hour
Answer:
4.29×10⁵ C
Explanation:
From the question,
The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.
E = 1/2mv²+mgh+Fd.................... Equation 1
Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.
But,
d = vt.................... Equation 2
Where v = velocity, t = time.
Substitute equation 2 into equation 1
E = 1/2mv²+mgh+F(vt)................... Equation 3
Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²
Substitute into equation 1
E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)
E = 260260+1622390+49608000
E = 51490650 J
Using,
E = qV................. Equation 4
Where q = charge of the battery, V = Voltage.
make q the subject of the equation
q = E/V............... Equation 5
Given: E = 51490650 J, V = 120 V
Substitute into equation 5
q = 51490650/120
q = 429088.75 C
q = 4.29×10⁵ C
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the other causes the electric field strength between them to be 1.9×105 N/C
Answer:
r = 6.5*10^-3 m
Explanation:
I'm assuming you meant to ask the diameters of the disk, if so, here's it
Given
Quantity of charge on electron, Q = 1.4*10^9
Electric field strength, e = 1.9*10^5
q = Q * 1.6*10^-19
q = 2.24*10^-10
E = q/ε(0)A, making A the subject of formula, we have
A = q / [E * ε(0)], where
ε(0) = 8.85*10^-12
A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)
A = 2.24*10^-10 / 1.6815*10^-6
A = 1.33*10^-4 m²
Remember A = πr²
1.33*10^-4 = 3.142 * r²
r² = 1.33*10^-4 / 3.142
r² = 4.23*10^-5
r = 6.5*10^-3 m
Use the following terms to write a short paragraph that describes what you learned from this lab: potential energy, kinetic energy, mass, velocity, collision and momentum. (underline or bold the terms in your paragraph, thank you!)
please help me~
due tomorrow at 12 pm
Answer:
Kinetic Energy is proportional to mass and velocity squared. Potential energy can be transferred to and from Kinetic Energy by doing work on a mass to raise it from the ground. Both momentum and kinetic energy are conserved during elastic collisions.
The lab illustrated energy transformation from potential to kinetic, the role of mass and velocity in kinetic energy and momentum, and the principles of elastic collision conserving total kinetic energy and momentum.
Explanation:Through this lab, I better understood the concepts of energy and its transformation. Initially, an object at rest on a height possesses potential energy, which depends on its mass and the height from the ground. As it falls, this potential energy converts into kinetic energy, which is directly proportional to the mass of the object and the square of its velocity. Moreover, when two objects collide, they exchange energy and momentum, where momentum is calculated as the product of an object's mass and velocity.
An elastic collision is a special scenario in which the total kinetic energy before and after the collision remains the same, demonstrating the law of conservation of momentum and kinetic energy.
An automobile tire having a temperature of 6.7 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 25 lb/in2 . What would be the gauge pressure in the tire when its temperature rises to 33◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and that the atmospheric pressure remains constant at 14.7 lb/in2 . Answer in units of lb/in2 .
Answer: The gauge pressure in the tire when its temperature rises to 33◦C will be 28.7 lb/in2
Explanation: Please see the attachments below
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air resistance is negligible, the distance from the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a.5m
b.10m
c.20m
d.40m
e.200m
Answer:
c.20
Explanation:
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k= 160 w/m·k). the fin diameter is d= 4.2 mm, and the fin is exposed to convective conditions characterized by h= 231 w/m2·k. it is reported that the fin efficiency is ηf= 0.65. determine (a) the fin length l and the fin effectiveness ɛf. account for tip convection.
Answer:
length of fin = 34.417
effectiveness = 33.77
Explanation:
the pictures attached below shows the whole explanation
During which phase of the moon may a solar eclipse occur?
Answer:
New moon
Explanation:
A solar eclipse can only take place at the phase of new moon, when the moon passes directly between the sun and Earth and its shadows fall upon Earth's surface.
A solar eclipse happens when the moon is perfectly aligned between the Earth and the Sun, thus obscuring it.
If the moon is between the Earth and the Sun, the far side of the moon is lighted, and thus you have a new moon.
Complete the sentence to explain when waves interact.
Waves interact with
and other
.
Answer:
Objects and waves
Explanation:
Answer:
Objects, Waves.
Explanation:
A wheel, starting from rest, rotates with a constant angular acceleration of 2.80 rad/s2. During a certain 5.00 s interval, it turns through 65.0 rad. (a) How long had the wheel been turning before the start of the 5.00 s interval? (b) What was the angular velocity of the wheel at the start of the 5.00 s interval?
Answer:
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s
Explanation:
Given that;
Initial Angular velocity = w1
Angular distance = s = 65 rad
time = t = 5 s
Angular acceleration a = 2.80 rad/s^2
Using the equation of motion;
s = w1t + (at^2)/2
w1 = (s-0.5(at^2))/t
Substituting the values;
w1 = (65 - (0.5×2.8×5^2))/5
w1 = 6rad/s
Time to reach w1 from rest;
w1 = at1
t1 = w1/a = 6/2.8 = 2.14s
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s
Answer:
a. The wheel was turning 2.14 s before the start of the 5.00 s interval.
b. The angular velocity of the wheel at the start of the 5.00 s interval was 6.00 rad/s.
Explanation:
At the start of the 5.00s interval, the wheel might have an initial angular velocity [tex]\omega_0[/tex]. We can obtain it from the kinematic equation:
[tex]\theta=\omega_0t+\frac{1}{2}\alpha t^{2}\\\\\implies \omega_0=\frac{\theta}{t}-\frac{1}{2}\alpha t[/tex]
Plugging in the known values for the time interval, the angular displacement and the angular acceleration, we get:
[tex]\omega_0=\frac{65.0rad}{5.00s}-\frac{1}{2}(2.80rad/s^{2})(5.00s)\\\\\omega_0=6.00rad/s[/tex]
It means that the angular velocity of the wheel at the start of the 5.00 s interval was 6.00 rad/s (b).
The time it took for the wheel to reach that angular velocity can be obtained from another kinematic equation:
[tex]\omega = \omega_0+\alpha t\\\\\implies t=\frac{\omega-\omega_0}{\alpha}[/tex]
It is important to take in account that in this case, the initial angular velocity is zero as the wheel started from rest, and the final angular velocity is the one we got in the previous question:
[tex]t=\frac{6.00rad/s-0}{2.80rad/s^{2}}\\\\t=2.14s[/tex]
Finally, the wheel was turning 2.14 s before the start of the 5.00 s interval (a).