At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius

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]

Answer:

A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

Answer:

Compound

Explanation:

A compound is a substance derived from the chemical combination of two or more elements

e.g Water ;

= [tex]H_2O\\Hydrogen\:and\:Oxygen[/tex]

Salt ;

[tex]NaCl\\Sodium\:and\: Chlorine[/tex]

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]

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

Answer:

1.8/0.61 =2.95 ft

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

;)

Answer:

The induced current is clockwise

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)

Answer:

433

Explanation:

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:

Answer:

Calcium has two electrons in its outer shell.

Explanation:

Calcium is defined as a metal due to its physical and chemical traits. The two outer electrons are very reactive. Calcium has a valence of 2.

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = [tex]q_{int}[/tex] / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

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.

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]

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:

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:

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

Answer:

A) E(r) = 1.3957 × 10^(5) N/C

B) E(r) = 9.8864 × 10⁴ N/C

C) E(r) = 1.13 × 10^(5) N/C

Explanation:

We are given;

q = 6 nc = 6 × 10^(-9) C

L = 10 cm = 0.1 m

d = 4.4 cm = 0.044 m

r1 = 1 cm = 0.01 m

r2 = 2 cm = 0.02 m

r3 = 3 cm = 0.03 m

Formula for the electric field strength in this question is given as;

E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)

When factorized, we have;

E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]

Plugging in the relevant values for q/(2π(ε_o)L)

We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m

Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53

Thus;

E(r) = 1078.52 [1/r + 1/(d - r)]

A) E1 is at r = 1 cm = 0.01m

Thus;

E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))

E(r) = 1.3957 × 10^(5) N/C

B) E2 is at r = 2 cm = 0.02 m

Thus;

E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))

E(r) = 9.8864 × 10⁴ N/C

C) E2 is at r = 3 cm = 0.03 m

Thus;

E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))

E(r) = 1.13 × 10^(5) N/C

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 .

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.

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m

Answer:

The polarity of the magnetic field changes

Explanation:

This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)

Answer:

3x10^5m/s

Explanation:

See attached file

Explanation:

The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.

Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.

[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]

Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.

[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]

[tex]v = 3\times 10^5\;\rm m/s[/tex]

Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].

To know more about the doppler effect, follow the link given below.

https://brainly.com/question/1330077.

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

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

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

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]

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.

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.

Answer:

A. Ef/ Ei = 1/2

B. EF/ Ei = 1

C Ef / Ei = 4

Explanation:

To solve this we apply Coulomb's law which States that

E = Kq / r^2

Where

q = charge r = straight line distance from q to the point in question and

K = Coulomb's constant

Then

Ei = K Q / r^2

So

A) If Q is halved then

Ef = K Q / (2 r^2)

Ef/Ei = 1/2

B) If R is halved, the value of the E-f

at a distance r remains unchanged. So

Ef/Ei = 1

C) if r is now r/2 then

Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2

Ef / Ei = 4