Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion of infinite charged sheets. This approximation is useful when a problem deals with points whose distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electric field from two finite sheets and compare it to the results for infinite sheets to get a better idea of when this approximation is valid. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2?

Given that,

Current, I = 400 A (downwards)

The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT.

We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current. We know that the magnetic force is given by :

[tex]F=iLB\sin\theta[/tex]

Here, [tex]\theta=90^{\circ}[/tex]

[tex]F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N[/tex]

The force is acting in a plane perpendicular to the current and the magnetic field i.e out of the plane.

Answer:

660V

Explanation:

V=IR

V=?,I=11A,R=60w

V=60×11

=660V

Answer:

(a)  Resulting torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

(b) Magnitude of resulting torque  =  = 3.99 Nm

(c) angular acceleration = = 0.009027 rad/s²

Explanation:

Given Data;

I = 442 kg˙m2

rx = 0.76 m,

ry = 0.035 m,

rz = 0.015 m,

Fx = 3.6 N,

Fy = -2.8 N,

Fz = 4.4 N

F = Fx i + Fy j + Fz ------------------------------1

r =  rx i + ry j + rz k ------------------------------2

(a) Torgue is given by the formula;

T = r * F  ------------------------------------3

Putting equation 1 and 2 into equation 3, we have;

Torque= r x F

            = (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )

            = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

Therefore,

Resulting torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

b)

Putting given values into the above expression, we have

 torque =  (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k

=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k

= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k

= (0.196) i - (3.29) j + (-2.254) k

Magnitude of resulting torque = √(0.196² + 3.29² +2.254²

                                                  =√15.943031

                                                  = 3.99 Nm

c) Angular acceleration is given by the formula;

angular acceleration = torque/moment of inertia

                                   = 3.99/ 442

                                  = 0.009027 rad/s²

Answer:

How much cash is expected for payments on account in June? = $412,400.

Explanation:

Cash Required for expected payments on account in June

Pilsner Inc. pays 25% on account in the month of billing and the remaining 75% in the next month, therefore, for the month of June the total payment will include 75% of Purchases made in May and 25% of purchases made in June.

Purchases in May = $411,200

Purchases in June = $416,000

Total Payment for May = (75% of 411,200)+(25% of 416,000)

= $308,400 + $104,000

= $412,400.

Final answer:

The multiverse concept along with the universe's infinite nature and lack of a center challenges our understanding, yet represents current scientific explanations supported by observational evidence and theoretical models.

Explanation:

The concept of the multiverse, suggesting that our universe is just one among countless others, is a challenging yet intriguing idea. The nature of cosmic expansion, driven by mass and dark energy, adds complexity to our universe's fate, with models predicting expansion forever or eventual contraction. While it may seem sensible to accept that the Big Bang led to a universe with no center or edge, and infinite potential, it's natural to find this overwhelming due to its departure from human intuition. These concepts push the boundaries of our understanding and stimulate philosophical and metaphysical discussions, demonstrating the dynamic involvement of spacetime. As opinions on what constitutes the 'center' or 'boundary' of an infinite universe vary, the scientific community continues to explore why the universe is structured as it is, why certain constants have their values, and what 'accidents' may have been necessary for existence as we perceive it.

The missing part of the question is;

1) How much heat is required to boil the water?

2) Assume that the liquid water takes up approximately zero volume, and the water vapor takes up some final volume Vf. You may also assume that the vapor is an ideal gas. How much work did the vapor do pushing on the piston?

3) How much did the water internal energy change?

Answer:

A) Q = 239.55 KJ

B) W = 18.238 KJ

C) ΔU = 221.31 J

Explanation:

We are giving;

Mass; m = 106 g

Latent heat of vaporization; L = 2260 J/g.

Molecular weight of water; M = 18 g/mol

Pressure; P = 101325 Pa

Temperature; T = 373.15 K

A) Formula for amount of heat required is;

Q = mL

Q = 106 x 2260

Q = 239560J = 239.55 KJ

B) number of moles; n = m/M

n = 106/18

n = 5.889 moles

Now, we know from ideal gas equation that;

PV = nRT

Thus making V the subject, we have; V = nRT/P

However, here we are told V is V_f. Thus, V_f = nRT/P

R is gas constant = 8.314 J/mol·K

Plugging in the relevant values ;

V_f = (5.889 x 8.314 x 373.15)/101325

V_f = 0.18 m³

Now, at constant pressure, work done is;

W = P(ΔV)

W = P(V_f - V_i)

W = 101325(0.18 - 0)

W = 101325 x 0.18

W = 18238.5J = 18.238 KJ

C) Change in water internal energy is gotten from;

ΔU = Q - W

Thus, ΔU = 239.55 KJ - 18.238 KJ

ΔU = 221.31 J

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    [tex]I_f_3 = 27.57 W/cm^2[/tex]

2 When third  polarizer is removed the intensity after it passes through the stack is    

                [tex]I_f_2 = 102.24 W/cm^2[/tex]

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  [tex]\theta_2 = 21^o[/tex]  

        The angle of the third  polarizing to the first is     [tex]\theta_3 = 61^o[/tex]

        The unpolarized light after it pass through the polarizing stack   [tex]I_u = 60 W/cm^2[/tex]

Let the initial intensity of the beam of light before polarization be [tex]I_p[/tex]

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     [tex]I_1 = \frac{I_p}{2}[/tex]

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    [tex]I_2 = I_1 cos^2 \theta_1[/tex]

                       [tex]= \frac{I_p}{2} cos ^2 \theta_1[/tex]

The intensity of light that will emerge from the third filter is mathematically represented as

                  [tex]I_3 = I_2 cos^2(\theta_2 - \theta_1 )[/tex]

                          [tex]I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)][/tex]

making [tex]I_p[/tex] the subject of the formula

                  [tex]I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}[/tex]

    Note that [tex]I_u = I_3[/tex] as [tex]I_3[/tex] is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      [tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}[/tex]

                      [tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}[/tex]

                           [tex]=234.622W/cm^2[/tex]

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      [tex]I_f_3 = \frac{I_p}{2} cos ^2 \theta_2[/tex]

[tex]I_f_3[/tex] is the intensity of the light emerging from the stack

substituting values

                     [tex]I_f_3 = \frac{234.622}{2} * cos^2(61)[/tex]

                       [tex]I_f_3 = 27.57 W/cm^2[/tex]

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  [tex]I_f_2 = \frac{I_p}{2} cos ^2 \theta_1[/tex]

[tex]I_f_2[/tex] is the intensity of the light emerging from the stack

Substituting values

                  [tex]I_f_2 = \frac{234.622}{2} cos^2 (21)[/tex]

                     [tex]I_f_2 = 102.24 W/cm^2[/tex]

The intensity of light after the second polarizer is removed is 22.88 w/cm² and after the third polarizer is removed is 26.73 w/cm².

The intensity of light passing through a polarizing filter can be determined by Malus's Law, which states that the intensity of the transmitted light is equal to the incident light multiplied by the square of the cosine of the angle between the filters. If unpolarized light is incident on the stack, the intensity is halved after passing through the first filter.

1) When the second polarizer is removed, the angle difference will be 40 degrees (61 degrees - 21 degrees). Hence, the intensity would be (60/2) * cos²(40 degree) = 22.88 w/cm².

2) When the third polarizer is removed, the angle difference is only 21 degrees, hence the intensity would be (60/2) * cos²(21 degree) = 26.73 w/cm².

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Answer:

12 m/s ∧2

Explanation:

The picture attached explains it all and i hope it helps. Thank you

To find the initial linear acceleration, calculate the initial torque due to gravity, the moment of inertia, and then apply Newton’s second law for rotation to obtain the initial angular acceleration. The initial linear acceleration can then be found by multiplying the initial angular acceleration by the length of the meter stick.

The question is asking for the initial linear acceleration of the end of the rigid body (meter stick) at the moment it is released.

The initial torque τ is equal to the gravitational forces acting on each particle times their respective distances from the pivot, summed up.

τ_initial = m*g*d1 + m*g*d2 = 0.4 kg * 9.8 m/s^2 * 0.6 m + 0.4 kg * 9.8 m/s^2 * 1.0 m

The moment of inertia I for the two-particle system can be calculated with the formula I = ∑mr^2 for each particle:

I = m*d1^2 + m*d2^2 = 0.4 kg * (0.6 m)^2 + 0.4 kg * (1.0 m)^2

According to Newton’s second law for rotation, the initial angular acceleration α is equal to the initial torque divided by the moment of inertia:

α = τ_initial / I

The initial linear acceleration a of the end of the body at the point opposite the pivot is equal to the product of the initial angular acceleration and the total length of the meter stick (1.0 m):

a = α * 1.0 m

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Answer:

1. A). Scalar

2.C) remains de same

Explanation:

Answer:

Explanation:

The pictures attached herewith shows the explanation and i hope it all helps. Thank you

Answer:

P = 7482600 Pa = 7.482 MPa

Explanation:

We have the equation:

P(V - Nb) = NKT

here,

P = Initial Pressure = ?

V = Initial Volume = 0.001 m³

N = No. of moles = 3

b = constant = 1.2 x 10⁻²⁸ m³

T = Temperature = 300 k

k = Gas Constant = 8.314 J/mol . k

Therefore,

P[0.001 m³ - (3)(1.2 x 10⁻²⁸ m³)] = (3 mol)(8.314 J/mol. k)(300 k)

P = (7482.6 J)/(0.001 m³)

P = 7482600 Pa = 7.482 MPa

Answer:

a) = 10.22 rad/s

b) = 0.35 m

Explanation:

Given

Mass of the particle, m = 1.1 kg

Force constant of the spring, k = 115 N/m

Distance at which the mass is released, d = 0.35 m

According to the differential equation of s Simple Harmonic Motion,

ω² = k / m, where

ω = angular frequency in rad/s

k = force constant in N/m

m = mass in kg

So,

ω² = 115 / 1.1

ω² = 104.55

ω = √104.55

ω = 10.22 rad/s

If y(0) = -0.35 m and we want our A to be positive, then suffice to say,

The value of coefficient A in meters is 0.35 m

Answer:

A spiral galaxy

Explanation:

Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

Explanation:

The magnitude of the electric potential difference is the induced emf and is given by the equation:

[tex]emf = l (v \times B)[/tex]

where l = length

v = velocity

B = magnetic field

As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. [tex]v \times B = vB sin90 = vB\\[/tex] ,

the induced emf = vlB, and thus increases.

The magnitude of the electric field potential difference between the wingtips increases

Answer:

The distance when the intensity is halved is 173.95 m

Explanation:

Given;

initial intensity of the sound, I₁ = 1.52 x 10⁻⁶ W/m²

initial distance from the explosion, d₁ = 123 m

final intensity of the sound, I₂ = ¹/₂ (1.52 x 10⁻⁶ W/m²) = 0.76 x 10⁻⁶ W/m²

Intensity of sound is inversely proportional to the square of distance between the source and the receiver.

I ∝ ¹/d²

I₁d₁² = I₂d²

(1.52 x 10⁻⁶)(123)² = (0.76 x 10⁻⁶)d₂²

d₂² = (1.52 x 10⁻⁶ x 123²) / (0.76 x 10⁻⁶)

d₂² = 30258

d₂ = √30258

d₂ = 173.95 m

Therefore, the distance when the intensity is halved is 173.95 m

The sound intensity varies inversely as the square of the distance from the

explosion source.

Reasons:

The sound intensity at 123 m = 1.52 × 10⁻⁶ W/m²

Required:

The distance at which the sound intensity is half the given value.

Solution:

Sound intensity is given by the formula;

[tex]I \propto \mathbf{ \dfrac{1}{d^2}}[/tex]

Which gives;

I × d² = Constant

I₁ × d₁² = I₂ × d₂²

Where;

I₁ = The sound intensity at d₁

I₂ = The sound intensity at d₂

[tex]d_2 = \mathbf{ \sqrt{\dfrac{I_1 \times d_1^2}{I_2} }}[/tex]

When the sound intensity is half the given value, we have;

I₂ = 0.5 × I₁

I₂ = 0.5 × 1.52 × 10⁻⁶ = 7.6 × 10⁻⁷

Therefore;

[tex]d_2 = \sqrt{\dfrac{1.52 \times 10^{-6} \times 123^2}{7.6 \times 10^{-7}} } \approx 173.95[/tex]

The distance from the explosion at which the sound intensity is half of the

sound intensity at 123 meters from the explosion, d₂ ≈ 173.95 m.

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Answer:

The final temperature is 50.8degrees celcius

Explanation:

Pls refer to attached handwritten document

Answer: 50.63° C

Explanation:

Given

Length of heater, L = 200 mm = 0.2 m

Diameter of heater, D = 15 mm = 0.015 m

Thermal conductivity, k = 5 W/m.K

Power of the heater, q = 25 W

Temperature of the block, = 35° C

T1 = T2 + (q/kS)

S can be gotten from the relationship

S = 2πL/In(4L/D)

On substituting we have

S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)

S = 1.2568 / In 53.33

S = 1.2568 / 3.98

S = 0.32 m

Proceeding to substitute into the main equation, we have

T1 = T2 + (q/kS)

T1 = 35 + (25 / 5 * 0.32)

T1 = 35 + (25 / 1.6)

T1 = 35 + 15.625

T1 = 50.63° C

Answer:

Around 2 hours

Explanation:

The answer is given in the pictures below. All the page numbers are circled on the top left corner of each page.

Final answer:

The question deals with calculating time based on the change in position of a reflected sunlight ray due to Earth's rotation, with the Earth rotating at a rate of 15 degrees per hour.

Explanation:

The question concerns the reflection of light from a mirror and the calculation of time based on the Earth's rotation rate. The sunlight's path changes throughout the morning due to the Earth's rotation, which moves at a consistent rate of 15 degrees per hour. Given the different positions where the ray of sunlight strikes the floor at different times, we can calculate the angle change and then the time elapsed.

To find the time elapsed, we first determine the angular change between the two observations. The change in position on the floor from 3.75 m to 1.18 m corresponds to a difference in angle when considering the mirror as the vertex of a right triangle with the wall. Next, we apply the Earth's rotational rate to find the corresponding time.

Answer:

2.8A

Explanation:

To calculate the total amplitude (when both pulses meet), we need to add up the amplitudes of each pulse. Since A is the amplitude of the first pulse, and we can call B the amplitude of the second pulse and C the total amplitude, we have that A+B=C=3.8A, which means that B=3.8A-A=2.8A, which we have already said is the amplitude of the second pulse.

Answer:

E = 3.194 x 10⁹ J = 3.194 GJ

Explanation:

The formula for the absolute potential energy is:

U = - GMm/2r

where,

G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

m = mass of satellite = 470 kg

r = distance between the center of Earth and satellite

Thus, the energy required from engine will be difference between the potential energies.

E = U₂ - U₁

E = - GMm/2r₂ - (- GMm/2r₁)

E = (GMm/2)(1/r₁ - 1/r₂)

where,

r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m

r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m

therefore,

E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)

E = 3.194 x 10⁹ J = 3.194 GJ

Answer:

100

Explanation:

[tex]\rho_m[/tex] = Resistivity of axon

r = Radius of axon

t = Thickness of the membrane

[tex]\rho_a[/tex] = Resistivity of the axoplasm

Speed of pulse is given by

[tex]v=\sqrt{\dfrac{\rho_mrt}{2\rho_a}}[/tex]

So, radius is given by

[tex]r=\dfrac{2\rho_a}{\rho_mt}v^2[/tex]

If radius is increased by a factor of 10 new radius will be

[tex]r_2=\dfrac{2\rho_a}{\rho_mt}(10v)^2\\\Rightarrow r_2=100\dfrac{2\rho_a}{\rho_mt}v^2\\\Rightarrow r_2=100r[/tex]

So, The radius will increase by a factor of 100.

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

Answer:

Explanation:

Let the length of inclined plane be L .

work done by gravity on the block

= force x length of path

= mg sinθ x L , m is mass of the block , θ is inclination of path

This in converted into potential energy of compressed spring

1/2 k x² = mgL sin31  , k is force constant . x is compression

.5 x 3400 x .37² = 33 x9.8 x sin31 L

L = 1.4

Length of incline = 1.4 m .

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, [tex]v=2.79\times 10^8\ m/s[/tex]

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

[tex]t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s[/tex]

Time period of oscillation measured by the observer is :

[tex]t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s[/tex]

So, the time period of oscillation measured by the observer is 5.79 seconds.

Answer:

1.)1.265+or minus 0.0006m

2).0.71%

Explanation:

See attached file

Answer:

increase, increase

Explanation:

If the bore of an engine is increased without any other changes except for the change to proper-size replacement pistons. the displacement will increase and the compression rate will increase.

Further Explanation:

If the bore is increased, there are several benefits to this. There is more area for the valves this improves flow through the engine. The compression ratio will increase, thereby increasing thermal efficiency. The displacement also increases as the power output increase. Also, higher RPM will further increase power.

However, a decrease in bore will result in corresponding reduction in displacement and power output.

Answer:

As they travel through rock, the waves move tiny rock particles back and forth -- pushing them apart and then back together -- in line with the direction the wave is traveling. These waves typically arrive at the surface as an abrupt thud. Secondary waves (also called shear waves, or S waves) are another type of body wave

Explanation:

Please mark me as brainliest!! Hope I helped

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

Answer:

0.012m

Explanation:

To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:

[tex]r=\frac{mv}{qB}[/tex]

m: mass of the proton 9.1*10^{-31}kg

q: charge of the proton 1.6*10^{-19}C

B: magnitude of the magnetic field 0.60T

v: velocity of the proton

In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:

[tex]qV=E_k=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.5*10^{3}V)}{1.67*10^{-27}kg}}=6.92*10^{5}\frac{m}{s}[/tex]

Then, by replacing in the formula for the radius you obtain:

[tex]r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m[/tex]

hence, the radius is 0.012m

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

Final answer:

To increase kinetic friction in curling, roughen the ice or stone's bottom; to decrease it, smooth the ice or use lubricants, often achieved by sweeping.

Explanation:

To increase the kinetic friction between the curling stone and the ice, thereby causing the stone to stop more quickly, you could roughen the surface of the ice or the bottom of the stone. This creates more irregularities which bump against each other, leading to increased vibrations and energy conversion to heat and sound, thus slowing down the stone. Additionally, players could sweep less or not at all in front of the stone so that the ice surface remains rougher and produces more resistance.

To decrease the kinetic friction and allow the stone to slide farther across the ice before coming to a stop, you could smooth the surface of the ice or apply a substance to the bottom of the stone to make it more slippery. Sweeping the ice in front of the stone is a common technique used to achieve this; it melts and smooths the ice surface, reducing the irregularities that cause friction and allowing the stone to glide more easily.

Answer: 120 kg

Explanation:

Given

Radius of balloon, r = 6.85 m

Distance moved by the balloon, d = 114 m

Time spent in moving, t = 17 s

Density of air, ρ = 1.2 kg/m³

Volume of the balloon = 4/3πr³

Volume = 4/3 * 3.142 * 6.85³

Volume = 4/3 * 3.142 * 321.42

Volume = 4/3 * 1009.90

Volume = 1346.20 m³

Density = mass / volume ->

Mass = Density * volume

Mass = 1.2 * 1346.2

Mass = 1615.44 kg

Velocity = distance / time

Velocity = 114 / 17

Velocity = 6.71 m/s

If it starts from rest, 0 m/s, then the final velocity is 13.4 m/s

acceleration = velocity / time

acceleration = 13.4 / 17 m/s²

The mass dropped from the balloon decreases Mb and increases buoyancy

F = ma

mg = (Mb - m) * a

9.8 * m = (1615.44 - m) * 13.4/17

9.8m * 17/13.4 = 1615.44 - m

12.43m = 1615.44 - m

12.43m + m = 1615.44

13.43m = 1615.44

m = 1615.44 / 13.43

m = 120.29 kg

Answer:

The magnetic field [tex]B_Z[/tex] [tex]= - 6.14*10^{-7} T[/tex]

Explanation:

From the question we are told that

      The wavelength is [tex]\lambda = 0.3m[/tex]

       The intensity is [tex]I = 45.0W/m^2[/tex]

       The time is [tex]t = 1.5ns = 1.5 *10^{-9}s[/tex]

Generally radiation intensity is mathematically represented as

              [tex]I = \frac{1}{2} c \epsilon_o E_o^2[/tex]

Where  c is the speed of light with a constant value of [tex]3.0 *10^8 m/s[/tex]

              [tex]E_i[/tex] is the electric field

             [tex]\epsilon_o[/tex] is the permittivity  of free space with a constant value of [tex]8.85*10^{-12} C^2 /N \cdot m^2[/tex]

 Making [tex]E_o[/tex] the subject of the formula we have

           [tex]E_i = \sqrt{\frac{2I}{c \epsilon_0} }[/tex]

      Substituting values

          [tex]E_i = \sqrt{\frac{2* 45 }{(3*10^8 * (8.85*10^{-12}) )} }[/tex]

               [tex]= 184.12 \ V/m[/tex]

Generally electric and magnetic field are related by the mathematical equation as follows

             [tex]\frac{E_i}{B_i} = c[/tex]

Where [tex]B_O[/tex] is the magnetic field

           making  [tex]B_O[/tex] the subject

                 [tex]B_i = \frac{E_i}{c}[/tex]

Substituting values

                 [tex]B_i = \frac{184.12}{3*10^8}[/tex]

                       [tex]= 6.14 *10^{-7}T[/tex]

Next is to obtain the wave number

  Generally  the wave number is mathematically represented as

                          [tex]n = \frac{2 \pi }{\lambda }[/tex]

Substituting values

                          [tex]n = \frac{2 \pi}{0.3}[/tex]

                              [tex]= 20.93 \ rad/m[/tex]

Next is to obtain the frequency

      Generally  the  frequency f is mathematically represented as

                    [tex]f = \frac{c}{\lambda}[/tex]

Substituting values

                   [tex]f = \frac{3 *10^8}{0.3}[/tex]

                      [tex]= 1*10^{9} s^{-1}[/tex]

Next is to obtain the angular velocity

                Generally  the  angular velocity  [tex]w[/tex] is mathematically represented as

                       [tex]w = 2 \pi f[/tex]

                           [tex]w = 2 \pi (1* 10^9)[/tex]

                              [tex]= 2 \pi * 10^9 rad/s[/tex]

Generally  the sinusoidal electromagnetic waves for the magnetic field B moving in the positive z direction is expressed as

                       [tex]B_z = B_i cos (nx -wt)[/tex]

Since the magnetic field is induced at the origin then the equation above is reduced to

                   [tex]B_z = B_i cos (n(0) -wt) = B_i cos ( -wt)[/tex]

x =0 because it is the origin we are considering

 Substituting values  

                      [tex]B_z = (6.14*10^{-7}) cos (- (2 \pi * 10^{9})(1.5 *10^{-9}))[/tex]

                           [tex]= - 6.14*10^{-7} T[/tex]