Answer:
Solar energy or infrared radiation energy
Provide 2 examples in sport or rehab where instantaneous velocity is more important than average velocity and 2 examples where average velocity is more important. Provide rationale.
Answer:
In high jump as well as long jump, instateneous velocity is more important than average velocity.
In relay races and 400m races, average is more important than instateneous velocity.
Explanation:
Instantaneous velocity is the velocity of anything in motion at a specific point in time. This is determined quite similarly to average velocity, however, we look at the period of time so that it approaches zero. If there is a standard velocity over a period of time, its average and instantaneous velocities may be the same. Instantaneous velocity is calculated as the limit as t approaches zero of the change in d over the change in t.
The range or length of long jump depends on the instantenous velocity of the jump and the height of high jump depends on the instantenous velocity of the height.
A person with greater average velocity wins a race. The average velocity of anything or object is referred to as its total displacement divided by the total time taken. That is to say, it is the rate at which an object changes its position from one place to another. Average velocity is also a Vector quantity. Meters per second is the SI unit. Although, any distance unit per any time unit can be used when necessary, such as miles per hour (mph) or kilometer per hour (kmph)
Instantaneous velocity is more important in specific instances like a tennis stroke or a football collision, while average velocity is more important for calculating average rates of change in situations like road trips or rehab.
Explanation:Instantaneous velocity is more important than average velocity in situations where we need to know the velocity at a specific instant in time. Examples of this include a tennis player's stroke in which they aim to hit the ball on the sweet spot of the racket for maximum velocity and minimal vibration, and in a collision in football where a player with the same velocity but greater mass has a greater impact due to their greater momentum.
On the other hand, average velocity is more important in situations where we need to calculate the average rate of change of position over a given time interval. Examples include calculating the average velocity of a car during a road trip to determine the time taken to reach a destination, and in rehab where the average velocity of a patient's movement is measured to track progress over time.
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Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small sphere, B with a net charge of -3.0 x 10^-6, when the distance between their centers is 10.0 meters k= 8.99 x10^9
Answer:
F = 5.33*10^-4N
Explanation:
to find the electrostatic force you use the Coulomb's law, given by the formula:
[tex]F=k\frac{q_Aq_B}{r^2}[/tex]
k: Coulomb's constant = 8.89*10^9 Nm^2/C^2
q_a: charge of A = 2.0*10^{-6}C
q_B: charge of B = -3.0*10^{-6}C
r: distance between the spheres = 10.0m
By replacing all these values you obtain:
[tex]F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N[/tex]
hence, the forcebetween the spheres is about 5.33*10^-4N
. If a substance has an excess number of electrons on its surface, what type of charge does it have?
A positive charge.
A negative charge.
A neutral charge.
A nuclear charge.
Answer:
A negative charge
Explanation:
If there is an excess of electrons in a substance, this will have a negative net charge. This is because electrons have negative charge
q = -1.67*10^{-19}
each electron will contribute with this charge to the total negative net charge of the substance.
hence, the answer is "A negative charge"
A substance with an excess number of electrons on its surface has a negative charge due to the negatively charged nature of electrons.
If a substance has an excess number of electrons on its surface, it will have a negative charge. This is because electrons have a negative charge, and having more electrons than protons results in a net negative charge for the substance. By contrast, if a substance had more protons than electrons, it would have a positive charge. Substances are typically electrically neutral when the number of electrons equals the number of protons, and this balance means there is no net electric charge.
An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Conversely, when an atom loses one or more electrons, it becomes a positively charged atom, also known as a cation. For instance, when a neutral sodium atom loses one electron, it becomes a sodium cation with a 1+ charge. Conversely, an oxygen atom that gains two electrons becomes an anion with a 2- charge.
Points Possible: 1, Points Correct: 0 Next If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed to listening for it standing by the track? In other words, how much faster will the sound travel through the iron rails than through the air? Assume the air temperature is 0°C. Sound will reach you seconds sooner.
Answer:
[tex]t_{air} - t_{iron} = 282.17 sec[/tex]
Explanation:
The distance covered = 100 km = 100,000 m
The speed of sound in iron, [tex]v_{iron}[/tex] = 5130 m/s
The speed of sound in air at 0° C, [tex]v_{air}[/tex]= 331.5°C
Speed = distance/time
Time = Distance /speed
[tex]t_{iron} = \frac{d_{iron} }{v_{iron} } \\t_{iron} = 100000/5130\\t_{iron} = 19.49 sec[/tex]
[tex]t_{air} = \frac{d_{air} }{v_{air} } \\t_{air} = 100000/331.5\\t_{air} =301.66 sec[/tex]
[tex]t_{air} - t_{iron} = 301.66 - 19.49\\ t_{air} - t_{iron} = 282.17 sec[/tex]
A worker pushes a wheelbarrow with a horizontal force of 50 N on level ground over a distance of 5.0 m. If a friction force of 43 N acts on the wheelbarrow in a direction opposite that of the worker, what work is done on the wheelbarrow by the worker?a) 250 J
b) 215 J
c) 35 J
d) 10 J
e) None of these answers is correct.
Answer:
Work done will be equal to 35 J
So option (c) will be correct answer
Explanation:
We have given force F = 50 N
Frictional force f = 43 N
So net force will be equal to [tex]F_{net}=50-43=7N[/tex]
Distance covered on the ground d = 5 m
We have to find the work done
Work done is equal to [tex]W=F_{net}\times d=7\times 5=35J[/tex]
So option (C) will be the correct answer.
The correct option is A 250 J.
Given, Horizontal force is 50 N.
Frictional force is 43 N.
The distance between worker and pushing of wheelbarrow is 5 m.
We know that, Work is a physics term describing the amount of energy transferred when it is moved over a distance by an external force.
So Work = Force . Displacement
work = F D cos [tex]\Theta[/tex]......(eq 1)
Here [tex]\Theta[/tex] is 0 .
So, [tex]work = 50 \times 5 \times 1[/tex] ( cos 0 = 1 )
Work = 250 J
Hence 250 J of work is done on the wheelbarrow by the worker.
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Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0 kg cart repels one way with a velocity of 0.12 m/s. The 2.0 kg cart goes the other way. At what velocity does it travel?
Answer:
V = -0.3 m/sec.
Explanation:
5.0 x 0.12 + 2.0 x v = 0. Which means that V = -0.3 m/sec.
The -ve sign shows it moves in the opposite direction.
You drop an irregular piece of metal into a container partially filled with water and measure that the water level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?
Answer:
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Explanation:
Given;
Mass m = 450g
Density = Mass/Volume = m/V
Volume V = change in height × base area = ∆h × A
∆h = 4.8cm
A = 8×8 = 64cm^2
V = 4.8×64 = 307.2cm^3
Density = 450g/307.2cm^3
Density = 1.4648g/cm^3
Density = 1.4648 × 1000kg/m^3
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Answer:
1464.84 kg/m³
Explanation:
Density = mass/volume.
D = m/v................. Equation 1
But from Archimedes principle,
every object immersed in water will displaced an amount of water equal to its own volume
Therefore,
v = v'................... Equation 2
Where v = volume of the irregular object, v' = volume of water displaced.
Since the base of the container is a square,
Then,
v' = L²(d)...................... Equation 3
Where L = length of the square base of the container, d = rise in water level.
Substitute equation 3 into equation 1
D = m/L²d......................... Equation 4
Given: m = 450 g = 0.45 kg, L = 8 cm = 0.08 m, d = 4.8 cm = 0.048 m
Substitute into equation 4
D = 0.45/(0.08²×0.048)
D = 0.45/0.0003072
D = 1464.84 kg/m³
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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the intensity at the central maximum of a double-slit interference pattern is 4I_1. The intensity at the first minimum is zero. at what fraction of the distance from the central maximum to the first minimum is the intensity I_1? Thank you.
The fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.
Explanation:The intensity at the central maximum of a double-slit interference pattern is 4I1 and the intensity at the first minimum is zero. To find the fraction of the distance from the central maximum to the first minimum, we can use the formula for intensity in double-slit interference:
I = 4I1 cos2(πd/λ)
At the first minimum, the intensity is zero. Therefore, we have:
0 = 4I1 cos2(πd/λ)
Simplifying this equation, we find that cos2(πd/λ) = 0.
If cos2(πd/λ) = 0, then cos(πd/λ) = 0.
And if cos(πd/λ) = 0, then πd/λ = π/2.
Solving for d/λ, we get d/λ = 1/2. Therefore, the fraction of the distance from the central maximum to the first minimum where the intensity is I1 is 1/2.
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Which situations contain unbalanced forces? Check all that apply.
Answer:
Explanation:
Without seeing the options i can tell you this:
A balanced force is equal on both sides
The friction would equal the force
The amount on the left (friction side) would equal the amount on right (force side)
Or the net force would equal zero
You should post the answer options for more help
Answer: A book falls to the floor.
A car skids to a stop.
A foam ball launches
Explanation:
A 2.05 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has force constant 830 N/m. The coefficient of kinetic friction between the floor and the block is 0.380. The block and spring are released from rest and the block slides along the floor.What is the speed of the block when it has moved a distance of 0.0200 m from its initial position?
Answer:
0.552 m/s
Explanation:
Given,
Mass of the block = 2.05 Kg
initial compression, x₁ = 0.03690 m
Spring constant, k = 830 N/m
coefficient of friction of the block, μ = 0.380
distance moved by the block,x₂ = 0.20 m
Speed, v = ?
Using conservation of energy
Initial spring energy + Work done by friction = Final spring energy + kinetic energy
[tex]\dfrac{1}{2}kx_1^2 - \mu mg x_2 = \dfrac{1}{2}kx_2^2 + \dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}\times 830\times 0.039^2 - 0.38\times 2.05\times 9.81\times 0.02 = \dfrac{1}{2}\times 830 \times 0.02^2 + \dfrac{1}{2}\times 2.05\times v^2[/tex]
v² = 0.3047
v = 0.552 m/s
Hence, the speed of the block is equal to 0.552 m/s
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
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).
In my solar system, we have a planet that is the innermost to our star that is exactly like the innermost planet in your solar system. The planet has a very large impact crater with a basin that covers a large region of the planet's surface, but few smaller craters have formed on top of it. What can one conclude from this?
Answer:
1. The planet doesn't have a thick enough atmosphere.
2. There have been multiple impacts on the planet.
Explanation:
As the planet is very close to the star, there is high possibility that it will not have an atmosphere. Just like Mercury doesn't have one. Presence of a very large crater with basin indicates that in the past a huge body had hit the planet and thus creating the crater with basin. Also, it must be very old.
Second observation that is given is the presence of smaller craters in the basin. This indicates impact craters created by smaller objects. If the planet had an atmosphere, these smaller objects would not be able to penetrate and reach the surface. Thus presence of these smaller crater indicate towards the planet not having any atmosphere.
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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If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?
The mass that should be displaced 20.0 cm should be 20000 N/m.
Calculation of the mass:At the time When a spring should be stretched or compressed its length so it changed by an amount x from its equilibrium length due to which the restoring force is exerted.
here,
spring constant is [tex]k = 1.00 * 10^3 N/m[/tex]
mass is x = 20.0 cm
Now
As per Hooke's law
We know that
F = - kx
[tex]= - 1.00 *10 ^3 * 20.0[/tex]
F = 20000 N/m
Hence, The mass that should be displaced 20.0 cm should be 20000 N/m.
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To find the restoring force, use Hooke's Law: F = -kx. With a spring constant of 1.00 × 10³ N/m and a displacement of 20.0 cm (0.200 m), the restoring force is 200 N.
To answer this, we use Hooke's Law, which states:
F = -kx
Where:
F indicates the restoring force
k denotes the spring constant
x indicates the displacement
Given the spring constant k = 1.00 × 10³ N/m and displacement x = 20.0 cm (which we convert to meters, so x = 0.200 m), we substitute these values into the equation:
F = - (1.00 × 10³ N/m) × (0.200 m)
This simplifies to:
F = -200 N
The negative sign indicates that the force is directed opposite to the direction of displacement, which is typical for restoring forces. Therefore, the restoring force is 200 N.
What are some of the phenotypic plant traits that scientists are investigating? Why do you think these traits were chosen?
The term phenotype in a more specific context to describe the collective expression of the genotype in conjunction with the environment on a plant's observable characteristics.
Explanation:
Phenotype, any of the measurable features of an organism arising from its genotype association (total genetic inheritance) with the environment. Behaviour, biochemical properties, colour, form, and scale are examples of measurable characteristics.An allele is another type of one gene. A phenotype triggered by the expression of a gene or group of genes is the external manifestation of a human trait. A trait is an attribute unique to an organism. The genotype's external expression is called the phénotype.Some of the phenotypic traits includes Height, color, shape.
What is a phenotype?Phenotype is a physical properties or expression of a gene in an individual.
Phenotype are features that is observable or can be seen. This could include the physical appearance where we have height, color, shape.
Phenotypic traits are the observable features such as plant height and eye color, hair color.
The genetic make up of an individual and environment determines the phenotype.Scientific investigation is based on the interest and expected output or result. For a trait or many traits to be investigated it must be related to the research to get the expected outcome.
Therefore, some of the phenotypic traits that scientist can investigate are height, shape and color.
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how many days are in a year on mars
Which is true about the potential energy stored in an inductor?
Answer:
It depends on the current
Explanation:
The formula for the energy stored in an inductor is W =1/2 *L*I2
We can state that an electric current flowing in inductor , there is energy stored in the magnetic field. let consider the pure inductor L, and the power which is supplied to initiate the current in inductor.
consider the equation P = i*v
write the equation in terms of inductor.
P = L*i di/dt
P*dt = L*i di
rewrite the Energy equation W = \int_{0}^{t}P*dt
W = \int_{0}^{t} L*i di
by solving we will get the W = 1/2 *L*I2
From this information and formula we can state that he energy stored in an inductor depends on the current.
Answer:
The complete option for the question
a.it depends on the self induced emf
b it depends on the inverse of the self induced emf
C.it depends on the current
d.it depends on the inverse of current
Answer is.it depends on the current
Explanation:
The energy stored in an inductor is W=1/2LI^2
From this equation it can be seen that energy in inductor L depends on current L
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.
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.
What bodie is outside the solar system
Answer:
Oumuamua is outside the solar system which is in cigar shaped.
Hope it will help you :))
Every star, except the Sun, is outside the solar system.
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]
Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction n2. If total internal reflection occurs at the boundary that separates medium 1 and medium 2, then we know which of the following? a) n1= n2 b) n1< n2 c) n1> n2 d) n1 ≥ n2
Answer:
option C
Explanation:
Given,
Refractive index of medium 1 = n₁
Refractive index of medium 2 = n₂
For total internal reflection to take place light should move from denser medium to the rarer medium.
Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2
n₁ > n₂
The correct answer is option C
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 pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s and that the radius of the loop-the-loop is 3.033 km. What is the apparent weight that the pilot feels (i.e., the force with which the pilot presses against the seat) at the bottom of the loop-the-loop?
Final answer:
The apparent weight that the pilot feels at the bottom of the loop-the-loop is calculated by summing the gravitational force and the centripetal force, resulting in an apparent weight of 4012.9 N.
Explanation:
The student's question is asking to calculate the apparent weight that a pilot feels at the bottom of a loop-the-loop. To find the apparent weight, we must calculate the normal force on the pilot, which is the combination of the gravitational force and the centripetal force required to keep the pilot in a circular path at the bottom of the loop.
Firstly, we calculate the gravitational force (weight) acting on the pilot: W = m × g, where m is the mass of the pilot, and g is the acceleration due to gravity. For the pilot with a mass of 84.0 kg, W = 84.0 kg × 9.8 m/s² = 823.2 N.
Next, we determine the centripetal force required for circular motion at the bottom of the loop: Fc = (mv²) / R, where m is the mass, v is the constant speed, and R is the radius of the loop. With v = 345 m/s and R = 3.033 km (or 3033 m), Fc = (84.0 kg × (345 m/s)²) / 3033 m = 3189.7 N.
The apparent weight of the pilot at the bottom of the loop is the sum of the gravitational force and the centripetal force: Apparent weight = W + Fc = 823.2 N + 3189.7 N = 4012.9 N.
Therefore, the pilot feels an apparent weight of 4012.9 N at the bottom of the loop-the-loop.
The layer of the sun that radiates most of the light that reaches earth is the
Answer:
photosphere
Explanation:
Photosphere, visible surface of the Sun, from which is emitted most of the Sun’s light that reaches Earth directly. Since the Sun is so far away, the edge of the photosphere appears sharp to the naked eye, but in reality the Sun has no surface, since it is too hot for matter to exist in anything but a plasma state—that is, as a gas composed of ionized atoms.
Source :
Photosphere astronomy
Written By:. Harold Zirin
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].
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 distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small compared to the distance from the slits to the screen, the distance between adjacent fringes near the center of the interference pattern _______.
Final answer:
The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment decreases by the same factor when the distance between the slits is decreased by a factor of 2.
Explanation:
The distance between adjacent fringes near the center of the interference pattern in a double-slit experiment is determined by the wavelength of the light and the distance between the slits. When the distance between the slits is decreased by a factor of 2, the distance between adjacent fringes near the center of the interference pattern also decreases by the same factor. This is because the interference pattern is directly related to the slit separation.