A heavier car is always safer in a crash than a lighter car.

Answer:

P₃₅₀ = 0.28 watt

Explanation:

First we find the resistance of the wire at 20°C:

R₀ = ρL/A

where,

ρ = resistivity = 1 x 10⁻⁶ Ωm

L = Length of wire = 100 cm = 1 m

A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²

Therefore,

R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)

R₀ = 1.27 Ω

Now, from Ohm's Law:

V = I₀R₀

where,

V = Potential Difference = ?

I₀ = Current Passing at 20°C = 0.5 A

Therefore,

V = (0.5 A)(1.27 Ω)

V = 0.64 volts

Now, we need to find the resistance at 350°C:

R₃₅₀ = R₀(1 + αΔT)

where,

R₃₅₀ = Resistance at 350°C = ?

α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹

ΔT = Difference in Temperature = 350°C - 20°C = 330°C

Therefore,

R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]

R₃₅₀ = 1.44 Ω

Now, for power at 350°C:

P₃₅₀ = VI₃₅₀

where,

P₃₅₀ = Power dissipation at 350°C = ?

V = constant potential difference = 0.64 volts

I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)

Therefore,

P₃₅₀ = V²/R₃₅₉

P₃₅₀ = (0.64 volts)²/(1.44 Ω)

P₃₅₀ = 0.28 watt

Answer:

2.95m

Explanation:

Using h= 2.5+ v²/2g

Where v= 3m/s

g= 9.8m/s²

h= 2.95m

Answer:

in the direction of the applied force

Explanation:

Answer:

The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.

Explanation:

The period of oscillation is calculated as;

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

where;

L is the length of the pendulum bob

g is acceleration due to gravity

If we make L the subject of the formula in the equation above, we will have;

[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]

The length of the pendulum depends on acceleration due to gravity (g).

Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.

Answer:

The effective (rms) current when the circuit is in resonance is 6 A

Explanation:

Given;

resistance of the resistor, R = 20 ohms

capacitance of the capacitor, C = 0.75 microfarads

inductance of the inductor, L =  0.12 H

effective rms voltage, [tex]V_{rms}[/tex] = 120

At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).

The effective (rms) current, = [tex]V_{rms}[/tex] / R

                                              = 120 / 20

                                              = 6 A

Therefore, the effective (rms) current when the circuit is in resonance is 6 A

Answer:

8.9 seconds

Explanation:

The height of the object at time t is:

y = h + vt − 4.9t²

where h is the initial height, and v is the initial velocity.

Given h = 30 and v = 40:

y = 30 + 40t − 4.9t²

When y = 0:

0 = 30 + 40t − 4.9t²

4.9t² − 40t − 30 = 0

Solving with quadratic formula:

t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)

t = [ 40 ± √(1600 + 588) ] / 9.8

t = 8.9

It takes 8.9 seconds for the object to land.

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

[tex]\omega=\omega_{0}+\alpha t[/tex]

Put the value in the equation

[tex]\omega=0.17+1.3\times1.7[/tex]

[tex]\omega=2.38(k)\ m/s[/tex]

We need to calculate the angular displacement

Using angular equation of motion

[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]

Put the value in the equation

[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]

[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]

[tex]\theta= 124.18^{\circ}[/tex]

We need to calculate the velocity at point A

Using equation of motion

[tex]v_{A}=v_{0}+\omega\times r[/tex]

Put the value into the formula

[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]

[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]

[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]

We need to calculate the acceleration at point A

Using equation of motion

[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]

Put the value in the equation

[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]

[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]

[tex]a_{A}=0.791j-0.851i[/tex]

[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]

Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]

(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]

Answer:

a) P = 10.27 kW

b) Pmax = 10.65 kW

c) E = 5.47 MJ

Explanation:

Mass of the loaded car, m = 950 kg

Angle of inclination of the shaft, θ = 28°

Acceleration due to gravity, g = 9.8 m/s²

The speed of the car, v = 2.35 m/s

Change in time, t = 14.0 s

a) The power that must be provided by the winch motor when the car is moving at constant speed.

P = Fv

The force exerted by the motor, F = mg sinθ

P = mgv sinθ

P = 950 * 9.8 *2.35* sin28°

P = 10,271.3 W

P = 10.27 kW

b) Maximum power that the motor must provide:

[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]

c) Total energy transferred:

Length of the track, d = 1250 m

[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

This question involves the concept of the law of conservation of momentum.

Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.

The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.

Mathematically,

[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]

where,

m₁ = mass of the first object

m₂ = mass of the second object

u₁ = velocity of the first object before the collision

u₂ = velocity of the second object before the collision

v₁ = velocity of the first object after the collision

v₂ = velocity of the second object after the collision

Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.

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The attached picture illustrates the law of conservation of momentum.

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

[tex] E = \frac{Kq}{d^{2}} [/tex]

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]

The electric field due to charge 2 is:

[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]

[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].

(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.

The answer can be explained as follows.

Given that the two charges are;

(a) At the midpoint; [tex]r = 0.5\,m[/tex].

We know that the electric field due to charge [tex]q_1[/tex].

Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]

The electric field due to charge [tex]q_2[/tex] is given by;

Therefore, the net electric field in the midpoint is given by;

The direction is towards the right side.

(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.

So the force on the charge is ;

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Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          [tex]Em_{f}[/tex] = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

[tex]E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201} \\\\E = 0.1108 \ N/C[/tex]

Therefore, the magnitude of the electric field is 0.1108 N/C

Answer:

Explanation:

The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .

Let the charge be - q .

force on charge

= q x E where E is electric field

= q x 680

weight = 1.6 x 10⁻³ x 9.8

so

q x 680 = 1.6 x 10⁻³ x 9.8

q = 1.6 x 10⁻³ x 9.8 / 680

= 23 x 10⁻⁶ C

- 23 μ C .

Answer:

[tex]F=1.8\times 10^{-3}\ N[/tex]

Explanation:

We have,

Length of wires is 11 m

Separation between wires is 0.033 m

Current in both the wires is 5.2 A

It is required to find the magnitude of force between two wires. The force between wires is given by :

[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]

So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]

Answer:

The correct answer is option 3 .

Please check the answer once :)

Answer:

The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.

Explanation:

The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?

Answer:

1.6 time constants must elapse

Explanation:

voltage on a cap, charging is given as

v = v₀[1–e^(–t/τ)]

Where R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

v = v₀[1–e^(–t/τ)]

1–e^(–t/τ) = 0.8

e^(–t/τ) = 0.2

–t/τ = –1.609

t = 1.609τ

Answer:

Majorly the electric field is reduced among other effect listed in the explanation

Explanation:

In capacitors the presence of di-electric materials

1. decreases the electric fields

2. increases the capacitance of the capacitors.

3. decreases the voltage hence limiting the flow of electric current.

 The di-electric material serves as an insulator between the metal plates of the capacitors

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

Diameter of [tex]25^{th}[/tex] Newton  Ring = 0.97 cm

Newton Rings is an experiment based on principle of  thin film interference

In Newton Rings Experiment the Diameter of  [tex]n^{th}[/tex] dark ring is given by equation (1)

[tex]\rm D_n= 2\sqrt{n\lambda R} ......(1)\\where \; \\D_n = Diameter\; of \; n^{th} \; dark \; ring }\\n = Number \; of \; Ring\\\lambda = Wavelength \\R = Radius \; of \; Curvature \; of\; the \; lens[/tex]

From the condition given

[tex]\rm D_5 = 0.3 \; cm \\D_{15} = 0.62 \; cm\\\\D_{25} = To \; be \; determined \\[/tex]

Putting the values in equation (1) for fifth diameter we get

[tex]\rm D_5 = 0.3=2\sqrt{5\lambda R}.......(2) \\D_{15} = 0.62 = 2\sqrt{15\lambda R}.......(3) \\\\Equation \; (3) - Equation (2) \\\\0.32 = 2\sqrt{\lambda R} ( \sqrt{15} -\sqrt{5})\\\\2\sqrt{\lambda R } = 0.1954....(4)\\[/tex]

So  From equation (1) and (4)

[tex]\rm Diameter \; of \; 25^{th} Ring =D_{25} = 2\sqrt{\lambda R } \times \sqrt{25} \\\\D_{25} = 0.1954\times 5 \\\\D_{25} = 0.97 \; cm[/tex]

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Answer:

f = 357.29Hz

Explanation:

In order to calculate the fundamental frequency in the closed tube, you use the following formula:

[tex]f_n=\frac{nv}{4L}[/tex]       (1)

n: order of the mode = 1

v: speed of sound = 343m/s

L: length of the tube = 24cm = 0.24m

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

[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]

The fundamental frequency of in the tube is 357.29Hz

0.17T

When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e

[tex]F_{C}[/tex] = [tex]F_{M}[/tex]     --------------(i)

But;

[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex]   [m = mass of the particle, r = radius of the path, v = velocity of the charge]

[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]

Substitute these into equation (i) as follows;

[tex]\frac{mv^2}{r}[/tex] = qvB

Make B subject of the formula;

B = [tex]\frac{mV}{qr}[/tex]            ---------------(ii)

Known constants

m = 1.67 x 10⁻²⁷kg

q = 1.6 x 10⁻¹⁹C

From the question;

v = 1.4 x 10⁷m/s

r = 0.85m

Substitute these values into equation(ii) as follows;

B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]

B = 0.17T

Therefore, the magnetic field strength is 0.17T

Answer:

D. All of the above

Explanation:

The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.

James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure,  moratorium, and achievement.

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

The ball reaches the maximum height of 54 feet

The question is about projectile motion,

the ball is shot at an angle α = 45°, and

the initial velocity u = 46 ft/s.

Under the projectile motion, the maximum height H is given by:

[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]

where, g = 9.8 m/s²

substituting the given values we get:

[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]

Hence, the maximum height is 54 feet.

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Answer:

The answer is A

Explanation:

Its A because he created a telescope to be able to observe stars.

Answer:

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Explanation:

Magnetic flux is the scalar product of the magnetic field over the area

               Ф = ∫ B. dA

where B is the magnetic field and A is the area

Let's look at stationary, for which factors affect flow

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Answer:

The  charge is  [tex]q = 3.14 *10^{-4} \ C[/tex]

Explanation:

From the question we are told that

     The mass of each ball is  [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]

       The distance of separation is  [tex]d = 1 \ m[/tex]

Generally the weight of the each ball is mathematically represented as  

      [tex]W = m * g[/tex]

where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]

substituting values

      [tex]W = 90.70 * 9.8[/tex]

      [tex]W = 889 \ N[/tex]

Generally  the electrostatic force between this balls is mathematically represented as

         [tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]

given that the the charges are equal we have

    [tex]q_1= q_2 = q[/tex]

So

         [tex]F_e = \frac{k * q^2 }{d^2}[/tex]

Now from the question we are told to find the charge when the weight of one  ball is equal to the electrostatic force

So  we have

       [tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]

   =>   [tex]q = 3.14 *10^{-4} \ C[/tex]

The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].

Given data:

The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.

The distance of separation of two balls is, d = 1 m.

First of all we need to obtain the weight of ball. The weight of the ball is expressed as,

W = mg

Here,

g is the gravitational acceleration.

Solving as,

W = 90.70 × 9.8

W = 888.86 N

The expression for the electrostatic force between this balls is mathematically represented as,

[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]

Since, the charges are equal then,

[tex]q_{1} =q_{2}=q[/tex]

Also, the magnitude of force between the balls is same as the weight of one ball. Then,

F = W

Solving as,

[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]

Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].

Learn more about the Coulomb's law here:

https://brainly.com/question/23202809

Answer:

the size are components relative to the whole.

Explanation:

they are particularly good at showing percentage or proportional data

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.