A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences

Answer:

B: I2>I1

Explanation:

See attached file

Answer:

6 N

Explanation:

From the laws of friction

F = ¶R = 0.3 × 20 = 6 N

The force of friction opposing the block's motion is 6 N.

The given parameters;

The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.

F = ma

Fₓ = μF

Substitute the given parameters to calculate the frictional force on the object.

Fₓ = 0.3 x 20

Fₓ = 6 N

Thus, the force of friction opposing the block's motion is 6 N.

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Answer:

The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily

Explanation:

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

Answer:

Length of a tube = 1.574 m

Explanation:

Given:

Fundamental frequency (f1) = 108 Hz

First overtone (f2) = 216 Hz

Speed of sound (v) = 340 m/s

Find:

Length of a tube

Computation:

We know that,

f = v / λ

f = nv / 2L  [n = number 1,2,3]

So,

f1 = 1(340) / 2L

f1 = 170 / L

L = 170 / 108 = 1.574 m

f2 = 2(340) / 2L

L = 340 / 216

L = 1.574 m

Answer:

t = 0.31s

Explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period  by using the information about the angular frequency:

[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]

Then the required time is:

[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]

Answer:

B = 0.025T

Explanation:

In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:

[tex]B=\frac{\mu N i}{L}[/tex]         (1)

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

N: turns of the solenoid = 500

i: current = 4.0A

L: length of the solenoid = 0.10m

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

[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]

The strength of the magnetic field at the center of the solenoid = 0.025T

Answer:

Magnetic field strength at the center is 2.51x10^-2T

Explanation:

Pls see attached file for step by step calculation

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])

[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]

Make sure your calculator is in degree mode.

-----------------

Method 2

[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]

-----------------

Method 3

[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

Answer:

The average current will be "17.69 nA".

Explanation:

The given values are:

Charge,

q = 9.2 pC

Time,

t = 0.52ms

The equivalent circuit of the cell surface is provided by:

⇒  [tex]i_{avg}=\frac{charge}{t}[/tex]

Or,

⇒  [tex]i_{avg}=\frac{q}{t}[/tex]

On substituting the given values, we get

⇒         [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]

⇒         [tex]=17.69^{-9}[/tex]

⇒         [tex]=17.69 \ nA[/tex]

Answer:

2.20°

Explanation:

For the central bright spot, we will use the constructive pattern for a double slit interference,

[tex]m\times w = d \times Sin\beta[/tex]

where w indicates the wavelength

and [tex]\beta[/tex] indicates the angle between the bright spot and center line.

now we will use the given values,

1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]

Solving for [tex]\beta[/tex],

[tex]\beta[/tex] = 2.204° ~ 2.20°

Therefore the correct answer is 2.20°

Answer:

Explanation:

The current through the  resistor is 0.83 A

.

Part b

The current through  resistor is 0.53 A

.

Part c

The current through  resistor is 0.30 A

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

[tex]K_{1} = K_{2} + W_{f}[/tex]

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.

[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]

Where:

[tex]f[/tex] - Friction force, measured in newtons.

[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.

[tex]m[/tex] - Mass of the toboggan, measured in kilograms.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]

If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:

[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]

[tex]f = 653.125\,N[/tex]

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]

[tex]\%K_{loss} = 92.889\,\%[/tex]

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]

[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]

[tex]\%v_{loss} = 73.333\,\%[/tex]

The speed of the toboggan is reduced in 73.333 %.

The average frictional force exerted on the toboggan by the rough surface is 661.5 N.

The percentage of the toboggan kinetic energy reduction is 7.11%.

The percentage of the toboggan speed reduction is 26.67%.

The given parameters;

The normal force on the toboggan is calculated as follows;

Fₙ = mg

Fₙ = 375 x 9.8 = 3675 N

The acceleration of the toboggan is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]

The coefficient of friction is calculated as follows;

[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]

The average frictional force exerted on the toboggan by the rough surface;

[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]

The initial kinetic energy of the toboggan is calculated as follows;

[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]

The final kinetic energy of the toboggan is calculated as follows;

[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]

The percentage of the toboggan kinetic energy reduction is calculated as follows;

[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]

The percentage of the toboggan speed reduction is calculated as follows;

[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]

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Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer:

Explanation:

Kinetic energy at the height = 1/2 m v²

= 1/2 x 750 x 20²

= 150000 J

Its potential energy = mgh

= 750 x 9.8 x 5

=36750 J

Total energy = 186750 J

Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy

If v be its velocity at the bottom

1/2 m v² = 186750

v = √498

= 22.31 m /s

Answer:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

Explanation:

The magnitude of force applied by each charge on one another can be given by Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

where,

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now, in the final state the charges on both spheres are halved. Therefore,

q₁' = q₁/2

q₂' = q₂/2

Hence, the new force will be:

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

using equation 1:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

The magnitude of the electrostatic force will be F' = F/4

Here we used Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

Here

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now

q₁' = q₁/2

q₂' = q₂/2

So, the new force should be

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

So,

F' = F/4

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Answer:

A, C

Explanation:

Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest

A. Replacing the string with a thicker string.

Thicker strings have more density. The more density the string has, the lower the sound.

Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:

For:

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

[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]

See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.

C. Doubling the length of the string.

Because the length of the string is inversely proportional to the frequency.

The longer the string, the lower the frequency.

So, if we double string, we'll hear lower sounds in any string instrument

--

In short,  for A, and C  We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Explanation: hope this helps ;)

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Answer:

v = 6t² − 2t^³/₂

s = 2t³ − ⅘t^⁵/₂ + 15

Explanation:

a = 12t − 3t^½

Integrate to find velocity.

v = ∫ a dt

v = ∫ (12t − 3t^½) dt

v = 6t² − 2t^³/₂ + C

Use initial condition to find C.

0 = 6(0)² − 2(0)^³/₂ + C

C = 0

v = 6t² − 2t^³/₂

Integrate to find position.

s = ∫ v dt

s = ∫ (6t² − 2t^³/₂) dt

s = 2t³ − ⅘t^⁵/₂ + C

Use initial condition to find C.

15 = 2(0)³ − ⅘(0)^⁵/₂ + C

15 = C

s = 2t³ − ⅘t^⁵/₂ + 15

Answer:

C. The atom loses 1 electron to have a total of 36.

Explanation:

Cations have a positive charge. Cations lose electrons.

The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

Answer:

From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.

Explanation:

The image is shown below.

Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.

Answer:

Explanation:

a. Is the sphere charged negatively or positively?

The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.

Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.

b. What is the magnitude of the electric field intensity between the plates?

The magnitude of the electric field intensity between the plates is 18400v/m.

C. Calculate the magnitude of the charge on the latex sphere.

The magnitude of the charge on the latex sphere hae been solved and attached

d. How many excess or deficit electrons does the sphere have?

There are 5 excess electrons that the sphere has.

Check the attachment for further explanation.

Answer:

The number of bright spot is  m =4

Explanation:

From the question we are told that

    The number of lines is  [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]

     The wavelength of the laser is  [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]

Now the the slit is mathematically evaluated as

        [tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]

Generally the diffraction grating is mathematically represented as

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

Here m is the order of fringes (bright fringes) and at maximum m  [tex]\theta = 90^o[/tex]

    So

          [tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]

=>        [tex]m = 4[/tex]

This  implies that the number of bright spot is  m =4

Answer:

W = 14700 J

Explanation:

This is an exercise on Newton's second law.

To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law

Y Axis . Perpendicular to the ramp

       N - Wy = 0

X axis. Parallel to the ramp, we assume it is positive when the ramp is going up

        F - Wx = m a

in this case F = 1470 N and it is parallel to the plane.

Work is defined by

      W = F .d

boldface indicates vectors

      W = F d cos θ

     let's calculate

      W = 1470 10 cos 0

       W = 14700 J

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex]        (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]

The strength of the magntetic field is 4.1mT