Four long wires are each carrying 6.0 A. The wires are located
at the 4 corners of a square with side length 9.0 cm. All of
these wires are carrying current out of the page. The
magnetic field (in T) at one corner of the square is:

Answers

Answer 1

Answer:

[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

[tex]B=\frac{\mu_oI}{2\pi r}[/tex]

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]

Then you have:

[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Four Long Wires Are Each Carrying 6.0 A. The Wires Are Locatedat The 4 Corners Of A Square With Side

Related Questions

16)

Gamma

rays

X-rays

UV

Infrared

Micro-

waves

Radio

waves

Visible light

Light is an electromagnetic wave and it has a place on the electromagnetic spectrum based on it energy and

wavelength. How does light's energy compare to the energy of other forms of electromagnetic waves?

A)

Light is less energetic than X-rays.

B)

Light is more energetic than X-rays.

Light is the least energetic electromagnetic wave.

D)

Light is the most energetic electromagnetic wave.

Answers

Answer:

Light is less energetic than X-rays.

Explanation:

The electromagnetic spectrum refers to the range of wavelengths or frequencies over which electromagnetic radiation extends. It is the entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light. In the electromagnetic spectrum, the entire distribution of electromagnetic radiation is done according to their frequency or wavelength.

The energy of an electromagnetic wave depends on its frequency and wavelength. The shorter the wavelength, the greater the energy of the electromagnetic wave but the larger frequency, the greater the energy of the electromagnetic wave.

X-rays has a frequency of about 1×10^20 Hz compared to visible light of frequency of about 1×10^15 Hz. Hence X-rays, having a larger frequency, is more energetic than visible light.

Identify the five categories of stressors.

Answers

Answer:

The five kinds of stressors are:

Acute time-limited

Brief naturalistic

Stressful events sequences

Chronic

Distant

Explanation:

yeah

A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)

Required:
a. Qualitatively, indicate the polarization of the plastic.

1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.

b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.

1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.

c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.

1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.

Answers

Answer:

(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center

(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.

Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.

(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.

Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,
creating a wave that moves through the crowd and they sit back down again after a while. This wave
move around the stadium without moving the people around it. Considering this situation, justify
your answer about nature of wave.

Answers

Answer:

The nature of the wave formed is a transverse  progressive wave.

Explanation:

A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.

A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.

The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.

A balloon with a radius of 16 cm has an electric charge of 4.25 10 –9 C.

Determine the electric field strength at a distance of 40.0 cm from the balloon’s centre.

Answers

Answer:

239 N/C

Explanation:

Electric field strength at distance R from a charge Q is given by the expression

E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .

R = 40 cm , Q = 4.25 x 10⁻⁹

Putting the given values

E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²

= 239 N/C .

Un levantador de pesas puede generar 3000 N de fuerza ¿Cuál es el peso máximo que puede levantar con una palanca que tiene un brazo de la fuerza de 2 m y un brazo de resistencia de 50 cm?

Answers

Responder: 12000N

Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%

Dado MA = Carga / Esfuerzo

Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)

MA = Carga / 3000

VR = 2 / 0.5 VR = 4

Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%

1 = (Carga / 3000) / 4

4 = Carga / 3000

Carga = 4 × 3000

Carga = 12000N

Esto significa que el peso máximo que se puede levantar es 12000N

9. How do air masses move?

Answers

Answer:

Air masses move with the global pattern of winds. In most of the United States, air masses generally move from west to east. They may move along with the jet stream in more complex and changing patterns. 

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume

Answers

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.

The given parameters;

diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³

The volume of the bobber is calculated as follows;

[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]

The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;

[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]

The volume of the lead is calculated as follows;

[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]

The buoyant force experienced by the lead due to the volume submerged is calculated as follows

[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]

The total buoyant force is calculated as;

[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]

The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;

[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]

Learn more here:https://brainly.com/question/17009786

Write the first equation of motion. Under what condition(s) is this equation valid?​

Answers

Explanation:

The first equation of motion in kinematics is given by :

[tex]v=u+at[/tex] .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?

Sort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.

mQ: the mass of the quarterback
mB: the mass of the football
(vQx)i: the horizontal velocity of quarterback before throwing the ball
(vBx)i: the horizontal velocity of football before being thrown
(vQx)f: the horizontal velocity of quarterback after throwing the ball
(vBx)f: the horizontal velocity of football after being thrown

Answers

Answer:

vBxf = 0.08625m/s

Explanation:

This is a problem about the momentum conservation law. The total momentum before equals the total momentum after.

[tex]p_f=p_i[/tex]

pf: final momentum

pi: initial momentum

The analysis of the momentum conservation is about a horizontal momentum (x axis). When the quarterback jumps straight up, his horizontal momentum is zero. Then, after the quarterback throw the ball the sum of the momentum of both quarterback and ball must be zero.

Then, you have:

[tex]m_Qv_{Qxi}+m_{Bxi}v_{Bxi}=m_Qv_{Qxf}+m_{Bxf}v_{Bxf}[/tex]    (1)

mQ: the mass of the quarterback = 80kg

mB: the mass of the football = 0.43kg

(vQx)i: the horizontal velocity of quarterback before throwing the ball = 0m/s

(vBx)i: the horizontal velocity of football before being thrown = 0m/s

(vQx)f: the horizontal velocity of quarterback after throwing the ball = ?

(vBx)f: the horizontal velocity of football after being thrown = 15 m/s

You replace the values of the variables in the equation (1), and you solve for (vBx)f:

[tex]0\ kgm/s=-(80kg)(v_{Bxf})+(0.46kg)(15m/s)\\\\v_{Bxf}=\frac{(0.46kg)(15m/s)}{80kg}=0.08625\frac{m}{s}[/tex]

Where you have taken the speed of the quarterback as negative because is in the negative direction of the x axis.

Hence, the speed of the quarterback after he throws the ball is 0.08625m/s

Two cars start moving from the same point. One travels south at 60 miyh and the other travels west at 25 miyh. At what rate is the distance between the cars increasing two hours later?

Answers

Answer:

65 m/h

Explanation:

Let the distance of the car moving south be y.

Let the distance of the car moving west be x.

Let the distance between the two cars be a.

These three distances can be represented as a right angled triangle. So we can say:

[tex]a^2 = x^2 + y ^2[/tex]

Let us differentiate with respect to time, since the distances are changing with respect to time:

[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)

da/dt = rate of change of distance between two cars

The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.

Therefore:

dy/dt = 60 m/h and dx/dt  = 25 m/h

After two hours, the distance of the two cars will be:

y = 2 * 60 = 120 miles

x = 2 * 25 = 50 miles

Therefore:

[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]

From (1):

130(da/dt) = 50(25) + 120(60)

130(da/dt) = 1250 + 7200 = 8450

da/dt = 8450/130 = 65 m/h

Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.

where would you expect to find vesicles of neurotransmitters
A. Synaptic gap
B. postsynaptic dendrites
C. Channels in the postsynaptic
D. Presynaptic terminal button

Answers

Answer:

D. Presynaptic terminal button

explanation:

Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron

hope this helped!

first law of equilibrium

Answers

Answer:

for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.

Hope I helped

Answer:

An object in static equilibrium has zero net force acting upon it.

The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as

 F = F1+ F2+ F3+ F4+. . . = 0

An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate the current and voltage output when the detector is used in the photo-conductive and photovoltaic modes respectively. The reverse saturation current (Is) is 9 nA.

Answers

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

Someone plzzz helpppppp with this last question

Answers

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

Which statement BEST explains the relationship between voltage, current, and power?

A. If voltage increases and everything else remains constant, then power will increase.

B. If voltage increases and everything else remains constant, then power will decrease.

C. If current decreases and everything else remains constant, then power will increase.

D. Voltage and power are inversely related.

Answers

I think the answer is c.if current decreases and everything else remains constant,then power will increase

To move a large crate across a rough floor, you push on it with a force at an angle of 15 degrees below the horizontal. Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.49.

Answers

Answer:

663N

Explanation:

We need to find the force that will overcome the frictional force.

The angle of the normal force is 15°.

The mass of the crate is 32 kg

The coefficient of static friction is 0.49

Frictional force is given in terms of Normal force as:

F = μNcosθ

where μ = coefficient of static friction

N = normal force

θ = angle of normal force

Frictional force is given as:

F = mg

=>mg = μNcosθ

=> N = mg/(μcosθ)

N = (32 * 9.8) / (0.49 * cos15)

N= 313.6 / 0.473

N = 663 N

The force needed to cause the box to move must be 663N or greater.

A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.

Answers

Answer:

d. 1.00 m

Explanation:

In 1905, Einstein proposed special theory of relativity of light.

This theory had a number of consequences or results. One of them is called "Length Contraction".

According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.

But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.

In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:

d. 1.00 m

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?

Answers

Answer:

(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

[tex]I(t)=I_{max}sin(\omega t)[/tex]   (1)

(a) The induced emf is given by the following formula

[tex]emf_L=-L\frac{dI}{dt}[/tex]    (2)

You derivative the expression (1) in the expression (2):

[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

[tex]emf_L=-\omega LI_{max}[/tex]

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

At t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

The current through an inductor of inductance L can be calculated by

[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1  

(a) The induced emf can be calculated by  

[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2  

Derivative the equation (1) in the equation (2)

[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]

(b) At t=0 the current is zero,

 

(c) At t = 0 the emf is:

[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]  

Therefore, at t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

To know more about inductance,

https://brainly.com/question/17431388

Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?

Answers

Answer:

A. K = 0.546 eV

B. cooper and iron will not emit electrons

Explanation:

A. This is a problem about photoelectric effect. Then you have the following equation:

[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex]   (1)

K: kinetic energy of the ejected electron

Ф: Work function of the metal = 2.48eV

h: Planck constant = 4.136*10^{-15} eV.s

λ: wavelength of light = 410nm - 750nm

c: speed of light = 3*10^8 m/s

As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :

[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]

B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm

[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]

You compare the energies E1 and E2 with the work functions of the metals and you can conclude:

sodium = 2.3eV < E1

cesium = 2.1 eV < E1

cooper = 4.7eV > E1 (this metal will not emit electrons)

iron = 4.5eV > E1 (this metal will not emit electrons)

A parallel-plate capacitor in air has a plate separation of 1.30 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and dis-connected from the source. The capacitor is then immersed in distilled water. Determine a) the charge on the plates before and after immersion.b) the capacitance and potential difference after immersion.c) the change in energy of the capacitor.

Answers

Answer:

Explanation:

capacitance of air capacitor

C = ε₀ A /  d

ε₀ is permittivity of medium , A is plate area , d is distance between plate .

C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²

= 170.19 x 10⁻¹⁴ F .

charge on the capacitor when it is charged to  potential of 255 V

= CV , C is capacitance and V is potential

= 170.19 x 10⁻¹⁴  x 255

= 4.34 x 10⁻¹⁰ C .

After it is disconnected from the source , and it is immersed in water , charge on it remains the same .

So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.

b )

When it is immersed in water its capacity increases  k times where k is dielectric constant of water which is 80 .

capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴  F

= 13615.2  x 10⁻¹⁴ F .

= 1.36 x 10⁻¹⁰ F

potential difference = charge / capacitance

= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰

= 3.2 V

c )

Energy of capacitor = 1/2 C V²

Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴

=  55.33 x 10⁻⁹ J

Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²

= .7  x 10⁻⁹ J .

decrease of energy = 54.63 x 10⁻⁹ J .

Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.
Car A: 500 kg, 10 m/s,
Car B: 2000 kg, 5 m/s,
Car C: 500 kg, 20 m/s,
Car D: 1000 kg, 20 m/s,
Car E: 4000 kg, 5 m/s, and
Car F: 1000 kg, 10 m/s.
(a) Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.
(b) Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.

Answers

Answer:

a)Car E = Car D  > (Car F = Car B = Car C) > Car A

b)Car E = Car D  > (Car F = Car B = Car C) > Car A

Explanation:

Car A: mass = 500 kg; speed = 10 m/s

Car B: mass = 2000 kg;speed = 5 m/s

Car C:mass = 500 kg; speed = 20 m/s

Car D: mass = 1000 kg; speed = 20 m/s

Car E:mass = 4000 kg; speed = 5 m/s

Car F: mass = 1000 kg; speed = 10 m/s

Part a) Now we know that momentum of each car is product of mass and velocity , so we will have

CarA:

[tex]P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s[/tex]

Car B:

[tex]P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s[/tex]

Car C:

[tex]P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s[/tex]

Car D:

[tex]P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s[/tex]

Car E:

[tex]P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s[/tex]

Car F:

[tex]P_6 = m v\\P_6 = (1000)(10)\\P_6 = 10^4 kg m/s[/tex]

So the momentum is given as ,

Car E = Car D  > (Car F = Car B = Car C) > Car A

Part b)Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car

so the order of impulse from largest to least is given as

Car E = Car D  > (Car F = Car B = Car C) > Car A

A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.

Answers

Answer:

The average  emf induce is   [tex]V = 2.625 * 10^{-5} \ V[/tex]

Explanation:

From the question we are told that

  The radius of the coil is  [tex]r = 0.30 \ m[/tex]

   The number of turns is  [tex]N = 420 \ turns[/tex]

    The frequency of the transition radio wave is  [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]

     The magnetic field is  [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]

     The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]

     

The period of the transmitted radio wave is  [tex]T = \frac{1}{f}[/tex]

    So  

              [tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]

 The potential difference can be mathematically represented as

               [tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]

           [tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]

Where  [tex]B_{min} = 0T[/tex]

substituting values

                   [tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]

                  [tex]V = 2.625 * 10^{-5} \ V[/tex]

Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m

Answers

Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]

Calculate the inlet and exit velocity of water jet

[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]

The conservation of mass equation of steady flow

[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]

[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]

[tex]A_e\ \texttt {is the exit area of the jet}[/tex]

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]

(b)

[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]

[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]

[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

Therefore, the horizontal force is -7603N

A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of the tooth on the prey, in N/cm², is
a) 0.0013 N/cm²
b) 128 N/cm²
c) 320 N/cm²
d) 640 N/cm²

Answers

Answer:

640N/cm^2

Answer D is correct

Explanation:

[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]

hope this helps

brainliest appreciated

good luck! have a nice day!

A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.

Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]

where

d = [tex]3.95 \times 10^{-2}m[/tex]

[tex]\lambda = 400 \times 10^{-9}m[/tex]

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.


What does this example show?

Answers

Answer:

This example shows that alloys are stronger than either of it's parent materials by themselves.

Explanation:

Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.

This is because an alloy is a mixture of two or more elements, where at least one element is a metal.

Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.

A soccer player is benched for being late to the game. In a fit of anger, she drops her ball from the top of the Physics building. It falls 4.9 meters after 1.0 second has elapsed. How much farther does it fall in the next 2.0 seconds

Answers

Answer:

The distance is  [tex]S = 39.2 \ m[/tex]

Explanation:

From the question we are told that

    The distance covered after t = 1 s is  [tex]d = 4.9 \ m[/tex]

   

According to the equation of motion

      [tex]v^2 = u^2 + 2ad[/tex]

 Now  u  =  0 m/s  since before the drop the ball was at rest

     [tex]v^2 = 2ad[/tex]

here  [tex]a =g = 9.8 \ m/s^2[/tex]

    So

       [tex]v = 9.8 m/s[/tex]

Also from equation of motion we have that

     [tex]S = ut + \frac{1}{2} at^2[/tex]

Now at  t = 2 s , as given from the question

  Then  u =  v = 9.8 m/s

And

     [tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]

     [tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]

    [tex]S = 39.2 \ m[/tex]

     

In the Life Cycle of Stars diagram, what stage does letter J represent?
A.) white dwarf

B.) black dwarf

C.) black hole

D.) neutron star

Which letters in the Life Cycle of Stars diagram represent stars on the main sequence?

A.) F & I

B.) C & G

C.) A & E

D.) B & D

In the Life Cycle of Stars diagram, what stage does letter L represent?

A.) neutron star

B.) black hole

C.) white dwarf

D.) black dwarf

In the Life Cycle of Stars diagram, what stage does letter I represent?

A.) neutron star

B.) black dwarf

C.) black hole

D.) white dwarf

In the Life Cycle of Stars diagram, what does letter D represent?

A.) a high mass star

B.) a white dwarf

C.) a cool star

D.) a low mass star

In the Life Cycle of Stars diagram, what stage does letter C represent?

A.) nuclear fusion

B.) a supernova

C.) a planetary nebula

D.) protostar formation

Which letter in the Life Cycle of Stars diagram represents a star forming region of space?

A.) M

B.) H

C.) J

D.) G

Which letter in the Life Cycle of Stars diagram represents a planetary nebula?
Group of answer choices

A.) G

B.) H

C.) L

D.) M

Answers

ANSWER: num 1 is black hole

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?

Answers

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

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