What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

Answer:

27°

Explanation:

The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)

So theta = arcsin(0.45)

=27°

The angle between the wire and the magnetic field is 27°.

Since The magnetic force per meter on a wire is measured to be only 45 %

So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field

Therefore,

theta = arcsin(0.45)

=27°

Hence, The angle between the wire and the magnetic field is 27°.

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Answer:

72J

Explanation:

distance moved is equal to 3m.then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

The answer is 72J.

Distance moved is equal to 3m.

Then just substitute x with 3m.

Fx = (14(3) - 3.0(3)2)) N

Fx =(42-18)N

Fx =24N

W=Fx *S

W=24N*3m

W=72J

A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

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Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

Answer:

Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are

Example of Conduction:

Heating a metal

Example of Convection:

Sea Breeze

Example of Radiation:

Sun

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Answer:

Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.

There are three(3) different ways heat can transfer and they are:

a) Conduction (through direct contact).

b) Convection (through fluid movement).

c) Radiation (through electromagnetic waves).

Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).

2. Baking a pie in an oven(radiation).

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Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]

Explanation:

Given: Speed of red light = 700 nm

= [tex]700\times10^{-9}[/tex] m

[tex]= 7\times10^{-7}[/tex] m

Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]

Speed of light = [tex]3\times10^8[/tex] m

Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]

[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]

Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]

The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].

Answer:

The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

Explanation:

Given;

first object with mass, m₁ = 285 kg

second object with mass, m₂ = 585 kg

distance between the two objects, r = 4.3 m

The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m

Gravitational force between the first object and the 42 kg object;

[tex]F = \frac{GMm}{r^2}[/tex]

where;

G = 6.67 x 10⁻¹¹ Nm²kg⁻²

[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]

Gravitational force between the second object and the 42 kg object

[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]

Magnitude of net gravitational force exerted on 42kg object;

F = 3.545x 10⁻⁷ N  -  1.727 x 10⁻⁷ N

F = 1.818 x 10⁻⁷ N

Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

The velocity and force are required.

The speed of the racket is 8.7 m/s

The required force is 471.43 N.

[tex]m_1[/tex] = Mass of racket = 1000 g

[tex]m_2[/tex] = Mass of ball = 60 g

[tex]u_1[/tex] = Initial velocity of racket = 12 m/s

[tex]u_2[/tex] = Initial velocity of ball = -15 m/s

[tex]v_1[/tex] = Final velocity of racket

[tex]v_2[/tex] = Final velocity of ball = 40 m/s

[tex]\Delta t[/tex] = Time = 7 ms

The equation of the momentum will be

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]

Force is given by

[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]

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Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

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Answer:

The  angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]

Explanation:

From the question we are told that

      The  angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]

       The  angular displacement is  [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]

From the first equation of motion we can define the movement of the record as

      [tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]

Given that the record started from rest [tex]w_o = 0[/tex]

So

       [tex]4.713^2 = 2 * \alpha * 25.14[/tex]

        [tex]\alpha = 0.4418 \ rad /s^2[/tex]

Answer:

Explanation:

The relation between activity and number of radioactive atom in the sample is as follows

dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms

For the beginning period

dN₀ / dt = λ N₀

58.2 = λ N₀

similarly

41 = λ N

dividing

58.2 / 41 = N₀ / N

N = N₀ x .70446

formula of radioactive decay

[tex]N=N_0e^{-\lambda t }[/tex]

[tex].70446 =e^{-\lambda t }[/tex]

- λ t = ln .70446 =   - .35

t = .35 / λ

λ = .693 / half life

= .693 / 5715

= .00012126

t = .35 / .00012126

= 2886.36

= 2900 years ( rounding it in two significant figures )

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

Answer:

D

Explanation:

Gravity is the force of attraction between two objects.

Each object creates a gravitational field in wich every other object is affected by it.

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

Answer:

Explanation:

The double slit interference phonemene is described for the case of constructive interference

          d sin θ= m λ                   (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

The double slit interference phonemene is described for the case of constructive interference

          d sin θ = m lam                    (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

we substitute

      sin θ = y / L

we substitute in equation 1

         d y / L = m λ

         λ = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 mm = 0.44 10⁻³ m

  y = 5.5 mm = 5.5 10⁻³ m

  L = 4.2m

  m = 1

let's calculate

        λ = 0.44  10⁻³ 5.5 10⁻³ / (4.2 1)

        λ = 5.76190 10-7 m

let's reduce to num

  lam = 5.56190 10-7 m (109 nm / 1m)

  lam = 556,190 nmtea

we substitute

      without tea = y / L

we substitute in equation 1

         d y / L = m lam

         lam = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 me = 0.44 10-3 m

  y = 5.5 mm = 5.5 10-3

  L = 4.2m

  m = 1

let's calculate

        lam = 0.44 10⁻³  5.5 10⁻³ / (4.2 1)

        lam = 5.76190 10⁻⁷ m

let's reduce to num

  lam = 5.56190 10⁻⁷ m (109 nm / 1m)

  lam = 556,190 nm

Answer:

gₓ = 6.52 m/s²

Explanation:

The value of acceleration due to gravity on the surface of earth is given as:

g = GM/R²   -------------------- equation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

M = Mass of Earth

R = Radius of Earth

Now, for the alien planet:

gₓ = GMₓ/Rₓ²

where,

gₓ = acceleration due to gravity at the surface of alien planet

Mₓ = Mass of Alien Planet = 2.4 M

Rₓ = Radius of Alien Planet = 1.9 R

Therefore,

gₓ = G(2.4 M)/(1.9 R)²

gₓ = 0.66 GM/R²

using equation 1

gₓ = 0.66 g

gₓ = (0.66)(9.81 m/s²)

gₓ = 6.52 m/s²

Answer:

B.distance between lines increases

Answer:

A. Lines become narrower

Explanation:

I  got it right on my quiz!

I hope this helps!! :))

Answer:

Zack should direct his throw outward and toward the back of the car.

Explanation:

As the car is moving forward, the book will be thrown with a forward component. Therefore, throwing this book backwards at a constant speed would cancel the motion of the car, allowing the book to have a greater chance of ending on the driveway. I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum.

The solution is throw 3.

I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

The statement that best applies Newton’s laws of motion to explain the skydiver’s motion is that an upward force balances the downward force of gravity on the skydiver. Newton's 3rd law often applies to skydiving.

When gravity is not acting upon the skydivers they would continue moving in the direction the vehicle they jumped from was moving. If no air resistance takes place, then the skydivers would still accelerating at 9.8 m/s until they hit the ground.

The skydiver after leaving the aircraft will accelerates downwards due to the force of gravity usually as there is no air resistance acting in the upwards direction, and there is a resultant force acting downwards, the skydiver will accelerates towards the ground.

Therefore, I say a greater chance as Zack may not have the exact timings as to land the book in his driveway. That too he may not have thrown the book with the right momentum as the skydivers.

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Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g

Answer:

0.866 A

Explanation:

From the question,

P = I²R............................. Equation 1

Where P = power, I = maximum current, R = Resistance.

Make I the subject of the equation

I = √(P/R).................... Equation 2

Given: P = 15 W, R = 20 Ω

Substitute these values into equation 2

I = √(15/20)

I = √(0.75)

I = 0.866 A

Hence the maximum current that can flow safely through the appliance = 0.866 A

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

Answer:

please brainliest!!!

Explanation:

V1/√T1 =V2/√T2

V1 = 331m/s

T1 = 0°C = 273k

V2 = ?

T2 = 35°c = 308k

Answer:

The answer is 40 N for APX

Explanation:

Answer:

The longer the cord, the lower the illumination

Explanation:

The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.

Long wires have more electrical resistance than shorter ones.

Let us consider this formula:

Resistance =[tex]\frac{\rho L}{A}[/tex]

From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.

Answer:

The magnitude of the net field located at the center of the square is the same in every of arrangement of the charges.

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

0.46N

Explanation:

See attached file

Answer:

The weight of the rod is 32.87 N

Explanation:

Density of the rod = 7800 kg/m

length of the rod = 81.2 cm = 0.812 m

diameter of rod = 2.60 cm = 0.026 m

acceleration due to gravity = 9.80 m/s^2

The rod can be assumed to be a cylinder.

The volume of the rod can be calculated as that of a cylinder, and can be gotten as

V = [tex]\frac{\pi d^{2} l}{4}[/tex]

where d is the diameter of the rod

l is the length of the rod

V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3

We know that the mass of a substance is the density times the volume i.e

mass m = ρV

where ρ is the density of the rod

V is the volume of the rod

m = 4.3 x 10^-4 x 7800 = 3.354 kg

The weight of a substance is the mass times the acceleration due to gravity

W = mg

where g is the acceleration due to gravity g = 9.80 m/s^2

The weight of the rod W = 3.354 x 9.80 = 32.87 N