Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential

Answers

Answer 1

Answer:

1. the sphere of the radius a

Explanation:

Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.

So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .

With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r

(for r ≥ R ) .

On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.

So Since b > a , the sphere of radius a will have the higher potential.

Also recall Because E = 0 inside a conductor, the potential


Related Questions

What is the reason for the increase and decrease size of the moon and write down in a paragraph.

Answers

Answer:

The reason for the increase or decrease of the moon is due to the angular perception of the moon.

Explanation:

Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.

The zenith is the highest part of the sky and the horizon the lowest.

When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.

But when looking up at the sky as there is no reference point there will be a failure in the perception of size.

At TTT = 14 ∘C∘C, how long must an open organ pipe be to have a fundamental frequency of 262 HzHz ? The speed of sound in air is v≈(331+0.60T)m/sv≈(331+0.60T)m/s, where TT is the temperature in ∘C∘C.

Answers

Answer:

Length of pipe organ(L) = 0647 m (Approx)

Explanation:

Given:

Temperature (T) = 14°C

Fundamental frequency (F) = 262 Hz

Speed of sound (v) = 331 + 0.60(T) m/s

Find:

Length of pipe organ(L)

Computation:

Speed of sound (v) = 331 + 0.60(14) m/s

Speed of sound (v) = 339.4

Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)

Length of pipe organ(L) = 339.4 / 2 (262)

Length of pipe organ(L) = 0647 m (Approx)

HELP!!! 35 point question. answer at least 3 correctly. please include equations and how you did it

Answers

Answer:

9. (B) ¼ Mv²

10. (A) √(3gL)

11. 20 N

12. 5 m/s²

Explanation:

9. The rotational kinetic energy is:

RE = ½ Iω²

RE = ½ (½ MR²) (v/R)²

RE = ¼ Mv²

10. Energy is conserved.

Initial potential energy = rotational energy

mgh = ½ Iω²

Mg(L/2) = ½ (⅓ ML²) ω²

g(L/2) = ½ (⅓ L²) ω²

gL = ⅓ L² ω²

g = ⅓ L ω²

ω² = 3g / L

ω = √(3g / L)

The velocity of the top end is:

v = ωL

v = √(3gL)

11. Sum of torques about the hinge:

∑τ = Iα

-(Mg) (L/2) + (T) (r) = 0

T = MgL / (2r)

T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)

T = 20 N

12. Sum of forces on the block in the -y direction:

∑F = ma

mg − T = ma

Sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a / R)

T = ½ Ma

Substitute:

mg − ½ Ma = ma

mg = (m + ½ M) a

a = mg / (m + ½ M)

Plug in values:

a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))

a = 5 m/s²

What is the threshold velocity vthreshold(ethanol) for creating Cherenkov light from a charged particle as it travels through ethanol (which has an index of refraction of n

Answers

Explanation:

The velocity of light in a medium of refractive index [tex]n[tex] is given by,

[tex]v=\frac{c}{n}[/tex]

[tex]v \text { is the velocity of light in the medium }[/tex]

[tex]c \text { is speed of light in vacuum }[/tex]

The exact value of speed of light in vacuum is [tex]299792458 \mathrm{m} / \mathrm{s}[/tex].

For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,

[tex]v_{\text {Cherenkov }} \geq \frac{c}{n}[/tex]

[tex]v_{\text {threshod }}=\frac{c}{n}[/tex]

For water [tex]n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}[/tex]

[tex]=225407863 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

For ethanol [tex]n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}[/tex]

[tex]=220435630 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

Answer:

The answer is "2.2 × [tex]\bold{10^8}[/tex]".

Explanation:

In the given question the value of n is missing which can be defined as follows:

n= 1.36

The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:

know we will have to use an equation as follows:

Formula:      

(ethanol) or the vthreshold = [tex]\frac{c}{n}[/tex]

                                         [tex]= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8[/tex]

The water in vthreshold:

[tex]= 2.2 \times 10^8 \ \ \frac{m}{ s} \\\\[/tex]

Express the value in c, that is multiple, so, the value of vthreshold(water) is:

=(0.735) c

A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens

Answers

Answer:

The focal length of the lens is 2.5 cm

Explanation:

Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;

[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]

Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:

[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]

then, solving for the focal distance knowing that the object distance is 1.5 cm:

[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]

The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)

Answers

Answer:

Explanation:

Let f₀ be the frequency .

energy of photons having frequency of f₀

= hf₀ where h is plank's constant

for electron to get ejected , work function should be equal to energy of photon

hf₀ = 3.55

4.1357 x 10⁻¹⁵ x f₀ = 3.55

f₀ = 8.58 x 10¹⁴ Hz .

A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached

Answers

Answer:

The  displacement is  [tex]A_r = 6.071 \ cm[/tex]

Explanation:

From the question we are told that

   The initial displacement is [tex]A_o = 10 \ cm[/tex]

     The displacement at the end of first oscillation is  [tex]A_d = 9.05 \ cm[/tex]

     

Generally the damping constant of this damped oscillator is mathematically represented as  

           [tex]\eta = \frac{A_d}{A_o}[/tex]

substituting values

           [tex]\eta = \frac{9.05}{10}[/tex]

        [tex]\eta = 0.905[/tex]

The displacement after 4 more oscillation is mathematically represented as

       [tex]A_r = \eta^4 * A_d[/tex]

substituting values

      [tex]A_r = (0.905)^4 * (9.05)[/tex]

      [tex]A_r = 6.071 \ cm[/tex]

Answer:

Displacement reached is 6.0708 cm

Explanation:

Formula for damping Constant "C"

[tex]C^n=\frac{A_2}{A_1}[/tex]                  where n=1,2,3,........n

Where:

[tex]A_2[/tex] is the displacement after first oscillation    

[tex]A_1\\[/tex] is the initial Displacement

[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]

In our case, n=1.

[tex]C=\frac{9.05}{10}\\C=0.905[/tex]

After 4 more oscillation, n=4:

[tex]C^4=\frac{A_6}{A_2}[/tex]                                        

Where:

[tex]A_6[/tex] is the final Displacement after 4 more oscillations.

[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]

Displacement reached is 6.0708 cm

An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room

Answers

Answer:

28.3°C

Explanation:

Using

T(t) = (T(0) - 20)*(e^(-k*t)) + 20

for some positive number k, and some initial temperature T(0).

Boundary conditions:

T(4) = T(0) - 5 _______ (i)

T(8) = T(0) - 7 _______ (ii)

==> solving for T(0) and k :

(i):

(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>

(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5

(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5

5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))

5 = (T(0) - 20) * ( 1 - e^(-k*4) )

(ii):

(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7

(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7

7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))

7 = (T(0) - 20) * (1 - e^(-k*8))

In both results, subsitute x = e^(-4k) and C = (T(0) - 20)

(i): 5 = C * (1 - x)

(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)

Substitute C*(1-x) from (i) into (ii):

(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5

back into (i):

(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3

C = T(0) - 20 ==>

T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3

= 28.3°C

In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?

Answers

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

the charge density in an insulateed solid sphere of radius find the electric field at a distance of from the center of the solid

Answers

Answer:

Assuming the charged density in the insulated solid sphere of radius 3.1m is 8.8e-9, the electric field at 5.2 meters is 73.1256 [tex]i[/tex].

Explanation:

The electric charge linear density is equal to 8.8 x[tex]10^{-9}[/tex]

the radius of the sphere is 3.1m

The magnitude of the electric field at the radius of the sphere equal to 5.2 meters can be calculated with the formula ;

- E = λ / 4πε₀ [ r  / α ( α + r ) ] [tex]i[/tex]

Solution:

E =  8.8 x[tex]10^{-9}[/tex] / 4πε₀ [ 3.1/ 5.2( 5.2 + 3.1) ] [tex]i[/tex]

= 1018.0995 [0.07183] [tex]i[/tex]

=  73.1256 [tex]i[/tex]

Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.

Answers

the signal to go back could possibly take 1 yr or maybe more

What is the displacement of the object after 3 seconds?

Answers

Answer:

3 meters

Explanation:

an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate

Answers

Answer:

The electric field  can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.

Explanation:

Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.

An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. Answer the following question:

The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true?

a. for every object submerged partially or completely in a fluid
b. only for an object that is floating
c. only for an object that is fully submerged and is sinking.
d. for no object submerged in a fluid

Answers

Answer:

Answer:

A. for every object submerged partially or completely in a fluid

Explanation:

This is following Archimedes principle which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether FULLYor PARTIALLY submerged, is equal to the weight of the fluid that the body displaces.

Explanation:

Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits

Answers

Answer:

The values is  [tex]m_{max} = 8001 \ bright \ spots[/tex]

Explanation:

From the question we are told that

    The slit distance is  [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]

    The  wavelength is  [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          [tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]

For maximum number of spot [tex]\theta = 90^o[/tex]

So  

       [tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]

        [tex]n =4000[/tex]

Now for the both sides plus the central maxima  we have

      [tex]m_{max} = 2 * n + 1[/tex]

substituting values

       [tex]m_{max} = 2 * 4000 + 1[/tex]

       [tex]m_{max} = 8001 \ bright \ spots[/tex]

   

Complete each of the statements

A. Lines of force are lines used to represent ________ an ________ electric field


B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge

C. The electric field is uniform if at any point in the field its _________ and ________ is the same

D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation

Answers

Answer:

A:  magnitude and direction

B: Force that the field exerts on a test charge

C: its magnitude and direction is the same.

D: electrostatic machine

two rollers that are driven by a motor that generates a rotation

Explanation:

IMPORTANT ANSWER ALL 3 PLEASE!

Answers

Answer:

4. Liters

5. Celsius

6. Grams

Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

Answers

Answer:

T = 3.14 hours

Explanation:

We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.

We know that the radius of Mars is 3,389.5 km.

So, r = 1,787 + 3,389.5 = 5176.5 km

Using Kepler's law,

[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]

M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]

So,

[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]

or

T = 3.14 hours

So, the orbital period is 3.14 hours

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on the

Answers

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]

[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]

[tex]\alpha=\cos^{-1}(0.3846)[/tex]

[tex]\alpha=67.38^{\circ}[/tex]

We need to calculate the angle β

Using cosine law

[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]

[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]

[tex]\beta=\cos^{-1}(0.923)[/tex]

[tex]\beta=22.63^{\circ}[/tex]

We need to calculate the force on 130 cm side

Using formula of force

[tex]F_{130}=ILB\sin\theta[/tex]

[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]

[tex]F_{130}=0[/tex]

We need to calculate the force on 120 cm side

Using formula of force

[tex]F_{120}=ILB\sin\beta[/tex]

[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]

[tex]F_{120}=0.1385\ N[/tex]

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

[tex]F_{50}=ILB\sin\alpha[/tex]

[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]

[tex]F_{50}=0.1385\ N[/tex]

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.

b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.

c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.

Given the following data:

Current = 4.00 Amperes.Magnetic field strength = 75.0 mT = [tex]7.5 \times 20^{-3}\;T[/tex]Length = 130 cm to m = 1.3 mHypotenuse = 130 cmOpposite side = 120 cmAdjacent side = 50 cm

Let us assume the current is flowing in a counterclockwise direction in the right-angle triangle.

First of all, we would determine the angles by using cosine rule:

[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]

[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]

a. To the determine the magnitude of the magnetic force on the 130 cm side:

Mathematically, the force acting on a current in a magnetic field is given by the formula:

[tex]F = BILsin\theta[/tex]

Where:

B is the magnetic field strength.I is the current flowing through a conductor.L is the length of conductor.[tex]\theta[/tex] is the angle between a conductor and the magnetic field.

Substituting the given parameters into the formula, we have;

[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]

b. To the determine the magnitude of the magnetic force on the 120 cm side:

[tex]F_{120}=BILsin\beta[/tex]

[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]

c. To the determine the magnitude of the magnetic force on the 50 cm side:

[tex]F_{50}=BILsin\alpha[/tex]

[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]

Read more: https://brainly.com/question/13754413

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .

Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.

Answers

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

Compare the value for the inductor when the current was increasing vs decreasing. Which statement matches the expected results. The inductance should be the same regardless of whether the current is increasing or decreasing. The inductance should be greater while the current is increasing. The inductance should be greater while the current is decreasing.

Answers

Answer:

see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,

Explanation:

The express for inductance is

         [tex]E_{L}[/tex]= L dI / dt

         L = E_{L}  (di / dt)⁻¹

where L is the inductance, E_{L} the induced electromotive force, di/dt  the variation of the current as a function of time.

When analyzing this equation we see that the inductance depends on the variation with respect to time of the current, therefore it is independent, increase decreases,

Correct answer the inductance must be the same regardless of whether the current increases or decreases.

A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.

Answers

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

A5 kg box slides 3 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3 m / s?

Answers

Answer:

Coefficient of kinetic friction (Cof. KE) = 0.153

Explanation:

Given:

Mass of box (M) = 5 kg

Distance = 3 m

Initial speed (v) = 3 m/s

Find:

Coefficient of kinetic friction (Cof. KE)

Computation:

v² = u² + 2as

a = v² / 2s

a = 9 / 2(3)

a = 1.5 m/s²

Coefficient of kinetic friction (Cof. KE) = a / g

Coefficient of kinetic friction (Cof. KE) = 1.5 / 9.8

Coefficient of kinetic friction (Cof. KE) = 0.153

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.

Required:
How far away did the impact occur?

Answers

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

 

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

       

     

An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.

Answers

Maybe it is around 300

Answer:

433

Explanation:

A radar pulse returns 3.0 x 10-4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object

Answers

Answer:

The distance is  [tex]D = 45000 \ m[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 3.0 *10^{-4 } \ s[/tex]

   

Generally the speed of the radar is equal to the speed of light and this has a value  

      [tex]c = 3.0*10^{8} \ m /s[/tex]

Now the distance covered by the to and fro movement of the radar is mathematically evaluated as

      [tex]d = c * t[/tex]

=>    [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]

=>  [tex]d = 90000 \ m[/tex]

Therefore the distance between the radar antenna and the object is  

      [tex]D = \frac{d}{2}[/tex]

       [tex]D = \frac{ 90000}{2}[/tex]

      [tex]D = 45000 \ m[/tex]

The distance between the radar antenna and the object will be 45000 m.

What is a radar antenna?

A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.

The given data in the problem is;

t is the time=  3.0 x 10⁻⁴

d is the distance between the radar antenna and the object=?

c is the peed of light=3×10⁸ m/sec

The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as

[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]

As a result, the radar antenna's distance from the target is

[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]

Hence the distance between the radar antenna and the object will be 45000 m.

To learn more about the radar antena refer to the link;

https://brainly.com/question/24067190

Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.

Answers

Answer:

Explanation:

magnetic field due to an infinite current carrying conductor

B = k x 2I / r where k = 10⁻⁷  , I is current in conductor and r is distance from wire

putting the given data

B = 10⁻⁷ x 2 x 100 / 8

= 25 x 10⁻⁷ T .

B )

earth's magnetic field = .5 gauss

= .5 x 10⁻⁴ T

= 5 x 10⁻⁵ T

percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100

= 5 %

C )

ii.  No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.

A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).

Answers

Answer:

The speed of the rod is 2.169 m/s.

Explanation:

Given that,

Mass = 0.100 kg

Current = 15.0 A

Distance = 2 m

Length = 0.550 m

Kinetic friction = 0.120

(a). We need to calculate the magnetic field

Using relation of frictional force and magnetic force

[tex]F_{f}=F_{B}[/tex]

[tex]\mu mg=Bli[/tex]

[tex]B=\dfrac{\mu mg}{li}[/tex]

Where, l = length

i = current

m = mass

Put the value into the formula

[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]

[tex]B=0.01425\ T[/tex]

[tex]B=1.425\times10^{-2}\ T[/tex]

(b). If the friction between the rod and rail is reduced zero.

So, [tex]f_{f}=0[/tex]

We need to calculate the acceleration

Using formula of force

[tex]F_{net}=f_{f}+F_{B}[/tex]

[tex]F_{net}=0+Bil[/tex]

[tex]ma=Bil[/tex]

[tex]a=\dfrac{Bil}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]

[tex]a=1.176\ m/s^2[/tex]

We need to calculate the speed of the rod

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value into the formula

[tex]v^2=0+2\times1.176\times2[/tex]

[tex]v^2=\sqrt{4.704}\ m/s[/tex]

[tex]v=2.169\ m/s[/tex]

Hence, The speed of the rod is 2.169 m/s.

an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas

Answers

Answer:

Explanation:

Charge on an electron (q) = 1.6 * 10 ^ -19 C

Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec

We know that, Force exerted on moving particle moving through a magnetic field :

[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]

1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B

B =  0.08573 T

Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?

Answers

Answer:

The energy contained in a single pulse is 90,200 J.

Explanation:

Given;

power of the laser, P = 82 TW = 82 x 10¹² W

time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s

the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m

The energy contained in a single pulse is calculated as;

E = Pt

where;

P is the power of each laser

t is the time to generate the power

E = (82 x 10¹²)(1.1 x 10⁻⁹)

E = 90,200 J

Therefore, the energy contained in a single pulse is 90,200 J

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