Nowdothesameproblemwiththepivotatthe toes. A Ballet dancer puts all her weight on the toes of one foot. If her mass is 60 kg, what is the force that has to be exerted by her leg muscle to hold that pose? Assume the pivot is at the toes.

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

Answer:

K = 3.9 kJ

Explanation:

The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:

[tex] K_{T} = K_{f} - K_{i} [/tex]  

The initial kinetic energy is:

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]

Where m₁ is the mass of the object before the explosion and v₁ is its velocity

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]

Now, the final kinetic energy is:

[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]

Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces

Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex]   (1)          

By conservation of momentum we have:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]  

[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]

[tex] v_{2} = \frac{5}{4}v_{1} [/tex]     (2)

By entering (2) into (1) we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]  

Hence, the kinetic energy added is:

[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]  

Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.

I hope it helps you!

Answer:

-2.24 m/s²

Explanation:

Given:

v₀ = 20 mi/hr = 8.94 m/s

v = 0 m/s

t = 4.0 s

Find: a

v = v₀ + at

0 m/s = 8.94 m/s + a (4.0 s)

a = -2.24 m/s²

Answer:

61.6° west of South

Explanation:

The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.

Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side  and the hypotenuse is 8

cos(θ) = [adjacent/hypotenuse]

Cos θ = 3.8/8

Cos θ = 0.475

θ = cos^-1 (0.475)

θ = 61.6°

Therefore the angle is 61.6° west of South.

Answer:

The height of the tower is 8.96 m.

Explanation:

We have, a visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s.

It is required to find the height of the tower. Let it is l. The time period of a simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum, or height of tower

[tex]l=\dfrac{T^2g}{4\pi^2}\\\\l=\dfrac{(6.01)^2\times 9.8}{4\pi^2}\\\\l=8.96\ m[/tex]

So, the height of the tower is 8.96 m.

Answer:

0.938 seconds

Explanation:

For the ball thrown upwards, we use the formula below to solve it:

[tex]s = ut - \frac{1}{2}gt^2[/tex]

where s = distance moved

u = initial speed = 19.2 m/s

t = time taken

g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

Let x be the height at which both balls are level, this means that:

=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)

For the ball dropped downwards, we use the formula below:

[tex]s = ut + \frac{1}{2}gt^2[/tex]

u = 0 m/s

At the point where both balls are level:

s = 18 - x

=> [tex]18 - x = 0 + 4.9t^2[/tex]

=> [tex]x = 18 - 4.9t^2[/tex]__________(2)

Equating both (1) and (2):

[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]

They will be level after 0.938 seconds

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

Answer:

Final Temperature = 71 °C

Explanation:

In this case, the ideal gas equation is written as;

PV = mRT

Where;

P is pressure

V is volume

m is mass

R is gas constant

T is temperature

We will take the volume to be constant.

So, in the initial state, we have;

P1•V = m1•R•T1 - - - eq(1)

In the final state, we have;

P2•V = m2•R•T2 - - - - eq(2)

Combining eq (1) and eq(2),we have;

P1•m2•R•T2 = P2•m1•R•T1

Dividing both sides by R gives;

P1•m2•T2 = P2•m1•T1

Making T2 the subject gives;

T2 = (P2•m1•T1)/(P1•m2)

Now, we are given;

m1 = 2kg

m2 = ½*2 = 1kg

P1 = 4 atm

P2 = 2.2 atm

T1 = 40°C = 273 + 40 K = 313K

Plugging in this values into the T2 equation, we have;

T2 = (2.2 × 2 × 313)/(4 × 1)

T2 = 344 K

Converting to °C, we have;

T2 = 344 - 273 = 71 °C

Answer:-

-1 m/s^2

Explanation:

The average acceleration is given by dividing the change in velocity by change in time;

[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]

[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]

the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.

Answer:

The force constant is  [tex]k =1.316 *10^{7} \ N/m[/tex]

The energy stored in the spring is  [tex]E = 1.68 *10^{7} \ J[/tex]

Explanation:

From the question we are told that

   The mass of the object is  [tex]M = 4.8*10^{5} \ kg[/tex]

    The period is [tex]T = 1.2 \ s[/tex]

The period of the spring oscillation is  mathematically represented as

         [tex]T =2 \pi \sqrt{ \frac{M}{k}}[/tex]

where  k is the force constant

   So making k the subject

       [tex]k = \frac{4 \pi ^2 M }{T^2}[/tex]

substituting values

       [tex]k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}[/tex]

      [tex]k =1.316 *10^{7} \ N/m[/tex]

The energy stored in the spring is mathematically represented  as

       [tex]E = \frac{1}{2} k x^2[/tex]

Where x is the spring displacement which is given as

        [tex]x = 1.6 \ m[/tex]

substituting values

      [tex]E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2[/tex]

       [tex]E = 1.68 *10^{7} \ J[/tex]

Answer:

[tex]\large \boxed{42\, \mu \text{C}}$[/tex]

Explanation:

The formula for the force exerted between two charges is

[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

[tex]F=k\dfrac{q^{2}}{r^2}[/tex]

[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]

Answer:

Explanation:

1 ) Since it is a isochoric process , heat energy passed into gas

= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .

= 7.4 x 12.47 x ( 500 - 300 )

= 18455.6 J.

2 ) Since there is no change in volume , work done by the gas is constant.

3 ) from  , gas law equation

PV = nRT

P = nRT / V

= 7.4 x 8.3 x 500 / .74

= .415 x 10⁵ Pa.

4 ) Average kinetic energy  of gas molecules after attainment of final temperature

= 3/2 x R/ N x T

= 1.5 x 1.38  x 10⁻²³ x 500

= 1.035 x 10⁻²⁰ J

1/2 m v² = 1.035 x 10⁻²⁰

v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶

= 1.49 x 10⁶

v = 1.22 x 10³ m /s

5 )  In this process , pressure remains constant

gas is cooled from 500 to 300 K

heat will be withdrawn .

heat withdrawn

= n Cp dT

= 7.4 x 20.79 x 200

= 30769.2 J .

6 )

gas will have reduced volume due to cooling

reduced volume = .74 x 300 / 500

= .444 m³

change in volume

= .74 - .444

= .296 m³

work done on the gas

= P x dV

pressure x change in volume

= .415 x 10⁵ x .296

= 12284 J.

Answer:

[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Explanation:

Angular acceleration

[tex]\begin{aligned}

\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\

\omega_{i} &=0 \\

\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\

&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\

&=27.02 \mathrm{rad} / \mathrm{s} \\

\alpha &=\frac{(27.02-0)}{3.15} \\

&=8.57 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

a)Tangential acceleration

[tex]\begin{aligned}

a &=r \alpha \\

&=\frac{12}{2} \times 10^{-2} \times 8.57 \\

a &=0.51 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

This question involves the concepts of the equations of motion for angular motion.

The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".

First, we will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 3.05 s

Therefore,

[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]

Now, the tangential acceleration can be given as follows:

[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]

a = 0.532 m/s²

Learn more about the angular motion here:

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The attached picture shows the angular equations of motion.

Answer:

Explanation: please see attached file I attached the answer to your question.

The angle between the magnetic field and the wire's velocity is 33.2 degrees.

Since the potential difference = 71mv = 71 *10 ^-3 V

The length is 12 cm = 0.12m

The magnetic field i.e. B = 0.27T

The speed or v = 4 m/s

here we assume [tex]\theta[/tex] be the angle

So,

e = Bvl sin[tex]\theta[/tex]

So,

[tex]Sin\theta[/tex] = e/bvl

= 71*10^-3 / 0.27 *4*0.12

= 0.5478

= 33.2 degrees

Therefore, the angle should be 33.2 degrees

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Answer:

Loudness: decreases

Amplitude: decreases

Pitch: stays the same

Frequency: stays the same

Explanation:

1.

An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.

2.

Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.

3 and 4.

The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.

Hope this helps!

Answer:

Q = 3877 KJ

Explanation:

Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:

Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)

Q = Q₁ + Q₂   ----- equation (1)

Now, for Q₁:

Q₁ = m C ΔT

where,

m = Mass of Water = 1.5 kg

C = Specific Heat of Water = 4187 J/kg.°C

ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C

Therefore,

Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)

Q₁ = 490 x 10³ J =490 KJ

Now, for Q₂:

Q₂ = m H

where,

m = Mass of Water = 1.5 kg

H = Heat of Vaporization of Water = 2258 KJ/kg

Therefore,

Q₂ = (1.5 kg)(2258 KJ/kg)

Q₂ = 3387 KJ

Substituting the values in equation (1), we get:

Q = Q₁ + Q₂

Q = 490 KJ + 3387 KJ

Q = 3877 KJ

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.

Given:

radius f the circle would be = 1.7/2 = 0.85 m

This rotation exercise can be treated using the rotation kinematics.

Angular acceleration:

θ = w₀ t + ½ α t²

t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)

 =>  θ = 0 + ½ α t²

 => α = 2θ / t²

=>  α= 2 × 2π / 1.2²

 => α = 4π = 8.7266 rad / s²

Let's calculate the angular velocity:

=> w = wo + α t

=>  w = 0 + α t

=> w = 8.7266 × 1.2

=> w = 10.47192 rad / s

The relationship between linear and angular velocity is

=> r = d / 2

=> r = 1.7 / 2 = 0.85 m

=> v = w r

=> v = 10.47192 × 0.85  

=> v = 8.90 m / s

Thus, the correct speed would be - 8.90 m/s

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Answer:

a. i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Power generated is 3.6 W.

Explanation:

Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.

i.e      V = IR

i. conducting wire

ii high-pass and low-pass filters

iii. Cobra-4 Xpert-link

iii. voltage source

b. Given that; V = 12 V and R = 40 Ohm's.

P = IV

From Ohm's law, I = [tex]\frac{V}{R}[/tex]

So that;

P = [tex]\frac{V^{2} }{R}[/tex]

   = [tex]\frac{12^{2} }{40}[/tex]

  = [tex]\frac{144}{40}[/tex]

 = 3.6 W

The power is 3.6 W.

Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.

(A) The wheel has angular velocity at time [tex]t[/tex] according to

[tex]\omega=\omega_0+\alpha t[/tex]

so that after 2.50 s, the wheel will have attained an angular velocity of

[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]

(B) The angular displacement of the wheel is given by

[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]

After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to

[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒  Magnetic force = Copper wire's weight

So,

⇒   [tex]B\times I\times L=M\times g[/tex]

On putting the estimated values, we get

⇒  [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]

⇒  [tex]50B=69.03297\times 10^{-3}[/tex]

⇒  [tex]B=1.38\times 10^{-3} \ T[/tex]

(b)...

As we know,

⇒  [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]

⇒      [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]

⇒      [tex]=2240\times \pi \ 0.000001[/tex]

⇒      [tex]=7.037\times 10^{-3} \ kg[/tex]

Answer:

Its initial position was 471 m.

Explanation:

We have,

Final position of the object is 327 m

Displacement of the object is -144 m

It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,

[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]

So, its initial position was 471 m.

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

     ΔE Δt> h ’/ 2

     h’= h / 2π

to relate this to the masses let's use Einstein's relationship

      E = m c²

let's replace

     Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

      Δm Δt> h ’/ 2c²

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C

B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C

C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Find out more information about magnitude here:

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Explanation:

We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of the pendulum

g is acceleration due to gravity on which it is placed

It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.

Answer:

Work-done in quarter an hour = 3.5 × 10⁶ J

Explanation:

Given:

Force (F) = 280 KN = 280,000 N

Velocity (V) = 50 km / h

Time (t) = 1 / 4 = 0.25 hour

Find:

Work-done in quarter an hour

Computation:

⇒ Displacement = Velocity (V) × Time

⇒ Displacement = 50 × 0.25

⇒ Displacement = 12.5 km

⇒ Work-done = Force (F) × Displacement

⇒ Work-done in quarter an hour = 280,000 × 12.5

⇒ Work-done in quarter an hour = 3,500,000

Work-done in quarter an hour = 3.5 × 10⁶ J