1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Calculate the energy required. 3) Calculate the average power required.

Answer:

The electric field between the plates is 2173 N/C

Explanation:

Given that,

Distance = 0.150 cm

Suppose,The initial speed of the electron is [tex]7.05\times10^6\ m/s[/tex]. The capacitor is 2.00 cm long,

We need to calculate the time

Using formula of time

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{2.00\times10^{-2}}{7.05\times10^{6}}[/tex]

[tex]t=2.8\times10^{-9}\ s[/tex]

We need to calculate the acceleration

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]a=\dfrac{2s}{t^2}[/tex]

Put the value into the formula

[tex]a=\dfrac{2\times0.150\times10^{-2}}{(2.8\times10^{-9})^2}[/tex]

[tex]a=3.82\times10^{14}\ m/s^2[/tex]

We need to calculate the electric field between the plates

Using formula of electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

Put the value into the formula

[tex]E=\dfrac{9.1\times10^{-31}\times3.82\times10^{14}}{1.6\times10^{-19}}[/tex]

[tex]E=2173\ N/C[/tex]

Hence, The electric field between the plates is 2173 N/C

Answer:

The minimum friction coefficient required is 0.3

(friction coefficients have no units)

Explanation:

To find a force we need to know something about a mass, and we haven't been told the mass of the car. Let's just call it 'm' and leave it at that for the moment, because it will cancel out in the end.

The centripetal force is given by F = ma = mv2/r

We have values for the velocity and the radius, so:

Fcent=m×6 × 6/13.5 = 2. 667m N

The frictional force must be equal to or greater than this force in order for the car to successfully make it around the curve without sliding out.

The frictional force will be given by:

Ffrict = μFnorm

Where Fnorm is the normal force, equal to mg.

We can equate these two forces, the frictional force and the centripetal force:

Fcent = Ffrict

2.667m=μmg

We can cancel out a factor of m in both sides and rearrange to make μ the subject:

μ = 2.667g

Substituting in the value g=9.8 ms−2,

μ = 2.667/9.8 = 0.27

Approximately = 0.3

Explanation:

Below is an attachment containing the solution.

Explanation:

The number of subshell present are s, p, d, and f

The number of orbitals in each are as follows -

The number of orbitals can be calculated by the degeneracy,  2 l + 1 , where l denoted is angular momentum quantum number that determines the shape of an orbital. For s, p, d, and f the angular momentum quantum number is 0, 1, 2 3 respectively.

So, maximum number of f orbitals that are possible can be calculated as -

2 ×  l + 1  =  2  × 3  + 1  =  7

Answer:

The population mean are the same

Explanation:

Answer:

The population means are the same.

Explanation:

The hypotheses for a difference in two population means are similar to those for a difference difference two population proportions.

At null point, Ha=0

Let the mean population of the young one be u1

Let the mean population of the old one be u2.

Then, the difference between their mean population distance is given as

Ha=u2-u1

Since, Ha is null point, Ha=0

0=u2-u1

u2=u1

This shows that the mean population distance of the old is equal to the mean population distance of the young.

Therefore their mean population distance is the same

Since it is null alternative then, the population mean are the same.

We must sample the population using

1. Samples must be random to remove or minimize bias.

2. Sample must be representative of the populations in question.

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

Final answer:

The most appropriate biomaterials for artificial tendons are polymers, due to their high toughness, elasticity, and ability to mimic natural extracellular matrices.

Explanation:

For fabricating an artificial tendon, which requires the capacity to sustain substantial deformation at low forces and quickly return to its original dimensions after stress release, polymers would be the most appropriate class of biomaterials. Biopolymers like collagen are noted for high toughness and elasticity, vital properties for mimicking the natural structure and function of tendons. Modern bioengineering strategies also involve creating synthetic extracellular matrices (ECM) which utilize peptides and polymer conjugates to mimic natural ECM, providing both structural support and facilitating tissue integration and regeneration.

Metals and ceramics, while being strong and rigid, do not offer the same flexibility and elastomeric properties required by tissues such as tendons that undergo repeated and significant strain. Semi-flexible polymers and polymer gels on the other hand, have been engineered to possess a balance of strength and flexibility, similar to that of natural tendon tissue.

Explanation:

As we know that

time = distance/speed

The time used for firs half of the trip was

(1 mi)(12 mi/hour) = 1/12 hours = 5 minutes

The last half of the trip will took 10 minutes, 1/6 hour.

Speed = distance/ time

(1 mil) = (1/6h) = 6 mil/h

so the speed for last half of the trip was = 6mph

the average speed was

(2mil)(1/4 hour) = 8 mil/hour

So the ling's average speed was 8mph.

Radiative energy

Explanation:

The light from polaris travels through space in the form of energy is called  

Radiative energy.

This energy is transferred by means of electromagnetic radiation like X-rays, gamma rays, light, heat radiation and so on. It can travel through space in the form of radiation. For instance, we get the heat through the sun, that is located far away from the Earth by means of radiation. Through the electromagnetic waves,  sun's heat is transmitted and not  through any kind of solid medium, but by means of vacuum.

Answer:

The magnitude of the other force is 43.75 N.

Explanation:

Given that,

Mass of the object, m = 8.5 kg

Force 1, [tex]F_1=14\ N[/tex]

Acceleration of the object, [tex]a=3.5\ m/s^2[/tex]

To find,

The greatest possible magnitude of the other force.

Solution,

Let [tex]F_2[/tex] is the magnitude of other force that is acting on the object. For the greatest force, the two forces must be act in opposite direction such that :

[tex]F_2-F_1=ma[/tex]

[tex]F_2=ma+F_1[/tex]

[tex]F_2=14+8.5\times 3.5[/tex]

[tex]F_2=43.75\ N[/tex]

So, the magnitude of the other force is 43.75 N.

Answer:

Torque = 882Nm

Explanation:

Torque = Mg×distance

But plank's is pivoted ,therefore distance=3/2=1.5m

Mass of Nancy=60jg

Acceleration due to gravity, g=9.8m/s^2

Torque= 60×9.8×1.5

Torque= 882Nm

Answer:

Cart 1 = 4 m/s

Cart 2 = 4 m/s

Explanation:

See attachment

Answer:

Explanation:

The energy change during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the bonds of the products less the chemical energy stored in the bonds of the reactants.

Hence:

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction absorbed energy and it is endothermic.

Yes, a vibrating proton would produce an electromagnetic wave, as accelerating charges emit radiation. This principle is central to many technological and scientific applications, including radio transmissions and the study of galactic structures in astronomy.

Accelerating charges such as protons, when they vibrate, indeed produce electromagnetic waves. This effect is due to the fact that a changing electric field generates a magnetic field, and a changing magnetic field, in turn, generates an electric field. As a proton oscillates, it experiences acceleration and therefore can emit radiation. This principle is extensively used in various technologies, like radio transmission, where an alternating current in an antenna accelerates charges and creates electromagnetic waves.

The production and detection of electromagnetic waves are crucial in many fields, including communications and astronomy. Just like an electron, a proton is also a spin 1/2 particle with a magnetic moment and can emit radiation that can be detected, such as the 21-cm line in the hydrogen spectrum, which allows astronomers to map the spiral arms of galaxies.

Considering the mass and charge of particles, a vibrating proton can generate electromagnetic radiation, albeit at different frequencies compared to electrons due to their larger mass. This forms the basis of nuclear magnetic resonance (NMR) utilized in various scientific and medical applications.

Answer:

C

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

                                 t = v / a

                                 t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

                                 t = v / g

                                 t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

Answer:

[tex]\omega_f = 1 rad/s[/tex]

Explanation:

Given,

After two revolutions, angular speed = 0.5 rev/s

Angular speed after 8 revolution = ?

Using equation of circular motion

[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]

[tex]0.5^2 =0^2 + 2\times \alpha\times 2[/tex]

[tex]\alpha = 0.0625 rad/s^2[/tex]

Now, Angular speed after 8 revolution

[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]

[tex]\omega_f^2 =0^2 + 2\times 0.0625 \times 8[/tex]

[tex]\omega_f = 1 rad/s[/tex]

Hence, the angular speed after 8 revolution is equal to 1 rad/s.

The angular acceleration of the fan blades is 0.25 rev/s². Using this constant angular acceleration, the angular speed of the blades after 8 revolutions is calculated to be 2 rev/s.

To solve this problem, we use the equation of motion for rotational systems.

Δω = α * Δθ Where Δθ is the change in angle (in rev), Δω is the change in angular speed (in rev/s), and α is the angular acceleration (in rev/s²). Given that the blades have an angular speed of 0.5 rev/s after 2 revolutions, we can solve for α: α = Δω / Δθ = (0.5 rev/s) / (2 rev) = 0.25 rev/s²

Then, knowing this constant angular acceleration, we can find the angular speed after 8 revolutions: Δω = α * Δθ

Δω = (0.25 rev/s²) * (8 rev) = 2 rev/s

So after 8 revolutions, the angular speed of the ceiling fan is 2 rev/s.

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Answer 62.78mph

Explanation:

Answer:

The returning speed = 70 miles per hour (mph)

Explanation:

Answer:

a) 2.4×10^4N/C

b) 2.9 ×10^3V

Explanation:

Correct ststement: An electric force of 3.9×10^-15N acts on the electron if it is placed anywhere between the two plates.

a) The electric field magnitude is given by E= F/e

Where F = electric force

e= elementary charge carried by a single proton e= 1.6×10^-19C

E= (3.9×10^-15)/(1.6×10^-19)

E= 2.4×10^4NC

b) Ptential difference is given by:

Change in V= E×change in distance

Potential difference= (2.4×10^4)× (0.12)

Potential difgerence= 2.9×10^3V

Answer: The total energy created by the Alpha decay.

Explanation: The sum total of the kinetic energy of the alpha particle and the new nucleus is the total energy created by the alpha decay.

Consider the decay of a radioactive nuclide by the spontaneous emission from its nuclei of alpha particles. An alpha particle which is composed of two protons and two neutrons and has a charge of +2. With an appreciable mass and its ejection from the nuclide creates a certain amount of recoil energy in the nucleus. The total energy (Ex) created by alpha decay is therefore the sum of the kinetic energy of the particle, the recoil energy given to the new nucleus, and the total energy of any emitted gamma rays.

The sum of the kinetic energies of an alpha particle and the new nucleus after alpha decay equals the initial energy (Q-value) of the reaction, most of which is carried by the alpha particle due to mass differences.

The sum of the kinetic energies of the alpha particle and the new nucleus after an alpha decay can be determined by considering the conservation of energy and momentum. The energy released during the decay, known as the Q-value, is primarily carried away by the alpha particle due to its relatively smaller mass compared to the daughter nucleus.

The energy of the alpha particle can be measured experimentally, which then allows us to determine the energy of the new nucleus. According to conservation of energy, the sum of the kinetic energies of the alpha particle and the new nucleus is equal to the Q-value of the reaction. If the Q-value (Qa) is known to be 4.3 MeV, and assuming the potential energy is zero in the final state, this energy will be distributed between the alpha particle and the new nucleus.

Using the example values provided, if the kinetic energy of the new nucleus (KEnucleus) is calculated to be 23.3 eV, then the remainder of the Q-value minus the kinetic energy of the new nucleus will represent the kinetic energy of the alpha particle. Since the alpha particle carries away most of the kinetic energy, the sum of the kinetic energies will be very close to the initial Q-value.

Explanation:

The magnetic field about a straight length of current carrying wire is circular in shape. As we know that the electric current passing through the wire is producing magnetic field. One can find out the direction of the magnetic field lines by using right hand rule or a compass. If the conductor is a straight wire then the magnetic field lines will form a concentric circles around the wire. Put you thumb in the direction of the electric current in the wire then the fingers will curl to show the direction of magnetic field which thus form the circular loop.

Answer:

circular in shape.

Explanation:

This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in.

Because the magnetic field created by the electric current in the wire is changing directions around the wire, it will repel both poles of the magnet by bending away from the wire.

Answer:

Passed into the power grid for others to use the electricity

Explanation:

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner

When a home using solar panels produces surplus electricity and has no battery system, the excess power is typically fed back into the electricity grid. This process, known as net metering, can often result in a credit from the electricity company.

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually fed back into the grid. This feed-in process involves the excess electricity being transmitted to the local electricity network where it can be used by other homes or businesses. In many places, electricity companies provide a credit to the solar power producer for this extra electricity, which can lower the overall utility bill.

This concept is known as net metering which allows homeowners to offset the cost of power drawn from the utility grid by pushing their surplus electricity into the grid. However, it's crucial to note that this process might depend on the policies of the local utility or the regional grid operator.

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Answer:

The angular frequency of oscillation w = 21

Explanation:

To solve the question, we note that x(t) is the point in the motion of the object. Therefore to find the angular frequency of oscillation, we find the relationship between the angular velocity and time

The angular frequency, ω is a scalar quantity used to depict the rate of rotation per unit time

When there is a function in simple harmonic motion (SHM) with the following equation then ω is the angular frequency

x(t) = A·cos (2πft) = A·cos(ωt)  which is similar to 42.5*cos(21t)

then 21 = angular frequency

Final answer:

The problem about a parallel-plate capacitor requires the application of electromagnetism principles to determine changes in potential difference, initial and final stored energy, and the work required to separate the plates, all based on the capacitor's geometry and the effect of plate separation on capacitance.

Explanation:

To answer the student's question about the parallel-plate capacitor, we need to apply concepts from electromagnetism, specifically the relationships between charge, voltage, capacitance, and energy in capacitors. When the parallel plates of a capacitor are separated, the capacitance changes, but the charge remains the same since it is isolated after being disconnected from the battery. The potential difference between the plates changes as a result of the changing capacitance. The initial energy stored in the capacitor can be calculated using the formula U = 1/2 CV^2, and the final energy stored after increasing the plate separation can be calculated with the same formula but with the new capacitance value. The work done to separate the plates is equivalent to the change in stored energy, which can either be the work done by an external force or the work lost to the system.

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

[tex]F_G=\frac{Gm_1m_2}{r^2}[/tex] where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

[tex]F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}[/tex] = [tex]\frac{2}{4} \frac{Gm_1m_2}{r^2}[/tex]

Therefore the gravitational force is halved. That is it will always decrease

Answer:

By measuring the time taken for the stars line of sight velocity to cycle from peak to Peak, and by calculation using newtons version of Kepler's third law

Explanation:

The motion of orbiting planets using planet-hunting techniques can rely on doppler effect. The light from the stars they orbit, as seen from Earth. As the star moves back and forth, the Doppler shift causes a slight change in its apparent colour which can be detected using spectroscopy. The blue shift and the red shift.

The Doppler technique of blue shift and the red shift can be used to estimate the semi major axis of the planets orbit by

Measuring the time it takes for the stars line of sight velocity to cycle from peak to Peak, and using newtons version of Kepler's third law

Given Information:  

Radius of circular loop = r = 9.1 cm = 0.091 m  

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻²  Ω/m

Number of turns = N = 1

Required Information:  

Electrical energy dissipated = E = ?  

Answer:  

Electrical energy dissipated = 50.09x10⁻³ Joules  

Step-by-step explanation:  

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.

Answer:

T- carrier

Explanation:

The T-carriers are frequently used for trunking between switching centers in a telephone network. It makes use if the same twisted pair copper wire that analog trunks employs. One pair for transmitting and the other pair for receiving.

Final answer:

To find the initial speed of the car, we use the work-energy principle and the equation μk × g × d = ½ × v². By substituting the given coefficient of kinetic friction (0.29) and skid mark distance (23.74 meters), we calculate the initial speed to be approximately 11.68 m/s.

Explanation:

The question involves calculating the initial speed of a car at the moment the driver applied the brakes, which resulted in skid marks on the road. The length of the skid marks and the coefficient of kinetic friction (μk) between the tires and the road are known.

To calculate the initial speed of the car, we use the work-energy principle, which states that the work done by the friction force is equal to the change in kinetic energy of the car:

Work done by friction = Change in kinetic energy

μk × m × g × d = ½ × m × v²

Since mass (m) cancels out and gravitational acceleration (g) is a constant (9.81 m/s²), we can simplify the equation to:

μk × g × d = ½ × v²

Now we can solve for the initial velocity (v):

v = √(2 × μk × g × d)

Plugging in the given values, μk = 0.29 and d = 23.74 meters, we have:

v = √(2 × 0.29 × 9.81 m/s² × 23.74 m)

v ≈ √(2 × 0.29 × 9.81 × 23.74)

v ≈ √(136.4)

v ≈ 11.68 m/s (approximately)

Thus, the initial speed of the car when the brakes were applied was approximately 11.68 meters per second.

Answer:

Explanation:

The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".

There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.

Answer:

4,5

Explanation:

The resulting swing converts potential energy to kinetic energy and kinetic energy to potential energy when the swinging stops. This is line with the law of conservation of mechanical energy which deals with inter conversion of energy forms and energy not being able to get lost.

The conservation of momentum is most suited to the collision. This law states that when a collision occurs the initial momentum before and after the collision is the same(without any external force).

The fundamental frequency of a piano wire can be calculated using the formula f = sqrt(T/(4L^2*m)). Plugging in the given values, the fundamental frequency of the piano wire is 116.6 Hz.

The fundamental frequency of a piano wire can be calculated using the formula:

f = √(T/(4L2m))

Where f is the fundamental frequency, T is the tension, L is the length, and m is the mass per unit length.

Plug in the given values:

T = 2030 N

L = 1.3 m

m = 0.003 kg/m

f = √(2030/(4*1.32*0.003))

Solving this equation gives the fundamental frequency f = 116.6 Hz.