If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

The definition of that same given problem is outlined in the following section on the clarification.

Explanation:

The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.  

Presently take a gander at the energy stored in your condensers while charging is Q.

⇒  [tex]U =\frac{Qmax^2}{C}[/tex]

So conclude C doesn't change substantially as well as,

When,

⇒  [tex]Q=\frac{Qmax}{2}[/tex]

⇒  [tex]Q^2=\frac{Qmax^2}{4}[/tex]

And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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Answer:

The  total number of maxima produced is   [tex]m_T = 7[/tex] maxima

Explanation:

From the question we are told that

    The number of lines per cm is  [tex]n = 5000 \ lines/cm[/tex]

      The wavelength of the light is  [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]

Now the distance between the lines is mathematically evaluated as

           [tex]d = \frac{1}{n}[/tex]

substituting values  

          [tex]d = \frac{1}{5000}[/tex]

          [tex]d = \frac{1 *10^{-2}}{5000}[/tex]     N/B - this  statement convert it from  cm to m

         [tex]d = 2 *10^{ -6} \ m[/tex]

Generally the condition for diffraction i mathematically represented as  

          [tex]dsin(\theta ) = m \lambda[/tex]

at maximum  [tex]\theta = 90 ^o[/tex]

             [tex]d sin (90) = \lambda m[/tex]

here m is the  number of  maxima  

      Thus  making  m the subject we have

          [tex]m = \frac{d sin (90)}{ \lambda }[/tex]

So     [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]

          [tex]m = 3.2[/tex]

=>          m  =3  

  Now the total number of maxima would include the bright fringe(3) and  dark fringe (3) plus the central maxima (1)

Thus  

      [tex]m_T = 3 + 3 +1[/tex]

       [tex]m_T = 7[/tex] maxima

Answer:

The next resonance will be 150 Hz.

Explanation:

The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.

The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.

Frecuency of the standing sound wave modes in a open-closed tube is:

fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)

The next resonance is at 90 Hz. This means that it occurs when n = 3:

f₃=3*30 Hz= 90 Hz

This means that the next resonance occurs when n = 5:

f₅=5*30 Hz= 150 Hz

The next resonance will be 150 Hz.

Answer:

I hope it is correct ✌️

The  lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.

The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.

Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of  the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So,  The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.

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Answer:

a) ±7.08×10⁻⁷ C/m²

b) 1.77×10⁻⁷ C

Explanation:

For a conductor,

σ = ±Eε₀,

where σ is the charge density,

E is the electric field,

and ε₀ is the permittivity of space.

a)

σ = ±Eε₀

σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)

σ = ±7.08×10⁻⁷ C/m²

b)

σ = q/A

7.08×10⁻⁷ C/m² = q / (0.5 m)²

q = 1.77×10⁻⁷ C

Answer:

Horizontal component: [tex]F_x = 58\ N[/tex]

Vertical component: [tex]F_y = 33.5\ N[/tex]

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

[tex]F_x = F * cos(angle)[/tex]

[tex]F_x = 67 * cos(30)[/tex]

[tex]F_x = 58\ N[/tex]

For the vertical component, we have:

[tex]F_y = F * sin(angle)[/tex]

[tex]F_y = 67 * sin(30)[/tex]

[tex]F_y = 33.5\ N[/tex]

So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.

Answer:

  F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

        F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

       F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

       F ’= k (q / 4 q / 4) / (0.5 d)²

       F ’= k q / 16 / 0.25 d²

       F ’= k q² / d²    0.0625 / 0.25

       F ’= F 0.25

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]

where,

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

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Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

(a) V = 65.625 Volts

(b) V = 131.25 Volts

(c) V = 131.25 Volts

Explanation:

Recall that:

1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:

[tex]V=k\frac{Q}{R}[/tex]

2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:

[tex]V=k\frac{Q}{r}[/tex]

where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.

Then, at a distance of:

(a) 48 cm = 0.48 m, the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]

(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:

[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

Answer:

c) a difference in electric potential

Explanation:

my insta: priscillamarquezz

Answer:

The atomic number

Explanation:

Answer:

h = 1.94 m

Explanation:

When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:

Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board

K.E = P.E

K.E = mgh

h = K.E/mg

where,

h = depth of well = ?

K.E = Kinetic Energy at bottom = 380 J

m = mass of bull dog and skate board = 20 kg

g = 9.8 m/s²

Therefore,

h = 380 J/(20 kg)(9.8 m/s²)

h = 1.94 m

Answer:

  a = 1.15 10³ g

Explanation:

For this exercise we will use the relations of the centripetal acceleration

     a = v² / r

where is the linear speed of the rotor and r is the radius of the rotor

let's use the relationships between the angular and linear variables

          v = w r

let's replace

          a = w² r

let's reduce the angular velocity to the SI system

        w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)

         w = 57.6 rad / s

let's calculate

       a = 57.6²  3.4

       a = 1.13 10⁴ m / s²

To calculate this value in relation to g, let's find the related

       a / g = 1.13 10⁴ / 9.8

       a = 1.15 10³ g

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

9,375

Explanation:

Data provided

The mass of the car m = 1500 Kg.

The diameter of the circular track D = 200 m.

For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-

The radius of the circular path is

[tex]R = \frac{D}{2}[/tex]

[tex]= \frac{200}{2}[/tex]

= 100 m

after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula

[tex]Force F = \frac{mv^2}{R}[/tex]

[tex]= \frac{1500\times (25)^2}{100}[/tex]

= 9,375

Therefore for computing the magnitude of the centripetal force we simply applied the above formula.

Answer:

clockwise

Explanation:

when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.

Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same  polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.

The direction of the induced current in the bottom ring is in the clockwise direction.

The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.

Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.

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Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

W = 12.96 J

Explanation:

Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.

A simple random sample is a sample drawn in such a way that each member of the population has equal chance for being included in the sample.

The mass percent of hydrogen in CH₄O is 12.5%.

Mass percent is the mass of the element divided by the mass of the compound or solute.

Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *

Therefore, The mass percent of hydrogen in CH₄O is 12.5%.

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Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

Increasing population

Explanation:

As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great

Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet  

Therefore the increasing population is the one and single most important reason

Answer:

t = 3.4 s

The box will come to rest in 3.4 s

Explanation:

For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma

Frictional Force = μR = μW = μmg

Therefore,

ma = μmg

a = μg

where,

a = acceleration of box = ?

μ = coefficient of sliding friction = 0.45

g = 9.8 m/s²

Therefore,

a = (0.45)(9.8 m/s²)

a = -4.41 m/s²  (negative sign due to deceleration)

Now, for the time to stop, we use first equation of motion:

Vf = Vi + at

where,

Vf = Final Speed = 0 m/s (since box stops at last)

Vi = Initial Speed = 15 m/s

t = time to stop = ?

Therefore,

0 m/s = 15 m/s + (-4.41 m/s²)t

(-15 m/s)/(-4.41 m/s²) = t

t = 3.4 s

The box will come to rest in 3.4 s

Answer:

b. only ray that reflects back in the same direction it came from

Explanation:

C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.

Correct answer is option B.

C-ray is the only ray that reflects back in the same direction it came from.

Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.

Answer:

I = 81.5 W/m^2

Explanation:

In order to calculate the intensity of the sound at the speaker, you use the following formula:

[tex]I=\frac{P}{A}[/tex]          (1)

P: power of the speaker's sound = 1.63W

A: surface area of the stereo set = 0.07m^2

You assume that the intensity of the sound at the speaker depends only of the surface area of the stereo set. Furthermore, you consider that the wave front of the sound is approximately plane.

You replace the values of the parameters in the equation (1):

[tex]I=\frac{1.63W}{0.02m^2}=81.5\frac{W}{m^2}[/tex]

The intensity of the speaker's sound at the speaker is 81.5 W/m^2

Answer:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

Answer:

Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second

Explanation:

Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg

Velocity of the car is 2.5 meter per second

Since formula to calculate the momentum of an object is,

p = mv

Where, p = momentum of the object

m = mass of the object

v = velocity of the object

By substituting these values in the formula,

p = [tex](6.3\times 10^{-3})\times 2.5[/tex]

  = [tex]1.575\times 10^{-2}[/tex] Kg meter per second

Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.