PHYSICS COLLEGE

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer 1

Answer:

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^(2) +2^(2) }$ = $√(13)$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^(2) + 3^(2) }$ = $√(25)$ = 5cm = 0.05m

electric potential V = $(kq)/(r)$

change in potential ΔV = $V_(1) - V_(2)$

ΔV = $(2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) )$ , where $q_(1) = q_(2)=$ 2.00μC

ΔV = $2kq((1)/(r_(1)) - (1)/(r_(2) ))$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $((1)/(0.036) - (1)/(0.05) )$

ΔV= 2.789×10⁵

$(1)/(2)mv^(2)$ = ΔV × q₃

$(1)/(2)$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

Answer 2

Answer:

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561

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How do you define 'heat' and 'temperature'

Answers

Temperature -a measure of the average kinetic energy of the individual particles in an object

Heat is the movement of thermal energy from a substance of higher temperature to a substance of lower temperature

Answer:

Heat and temperature are related but very different.

Explanation:

Heat: The total energy of molecular motion in a substance

Temperature: A measure of the average energy of molecular motion in a substance

For further help:

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A 200-lb object is released from rest 600 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is minus10v, where v is the velocity of the object in ft/sec, determine the equation of motion of the object. When will the object hit the ground? Assume that the acceleration due to gravity is 32 ft divided by sec squared and let x(t) represent the distance the object has fallen in t seconds.

Answers

Answer:

x(t) = 20t + 12.75e⁻¹•⁶ᵗ + 487.5

t = 24.375 s

Explanation:

The force balance on the object is given as

Net force = W - Drag force

ma = W - 10v

a = (dv/dt)

ma = m(dv/dt) = 200 - 10v

W = mg

200 = m×32

m = 6.25 kg

m(dv/dt) = 200 - 10v

6.25(dv/dt) = 200 - 10v

(dv/dt) = 32 - 1.6v

v' + 1.6v = 32

Solving this differential equation using the integrating factor method

(ve¹•⁶ᵗ) = ∫ (32e¹•⁶ᵗ) dt

ve¹•⁶ᵗ = (20e¹•⁶ᵗ) + c (where c = constant of integration)

v = (20 + ce⁻¹•⁶ᵗ)

At t = 0, v = 0

0 = 20 + c

c = -20

v = (20 - 20e⁻¹•⁶ᵗ)

v = (dx/dt)

(dx/dt) = 20 - 20e⁻¹•⁶ᵗ

dx = (20 - 20e⁻¹•⁶ᵗ) dt

x(t) = 20t + 12.5e⁻¹•⁶ᵗ + c (c is still the constant of integration)

At t = 0, x = - 500

- 500 = 0 + 12.5 + c

c = 512.5

x(t) = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

when the object hits the ground, x = 0

0 = 20t + 12.75e⁻¹•⁶ᵗ - 487.5

20t + 12.75e⁻¹•⁶ᵗ = 487.5

Solving by trial and error,

t = 24.375 s

Hope this Helps!!!

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of increasing photon energy, lowest first.

Answers

Answer:

Explanation:

1. radio waves from am

2. radio waves from fm

3.yellow light from a sodium street lamp

4. microwaves from an antenna of a communications system.

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