A light wave in space is described by the general function: u(r, t) = 1 (2π) 3/2 Z dk A(k) e i(k·r−ωt) . (1) (a) Find the particular solution knowing that at t = 0 the wave is given by

Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)

Explanation:

The quantity of heat is required to raise the temperature of of water is 94050 joule.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.

Mass of water: m = 450 grams

Initial temperature of water = 35° C

Final temperature of water = 85° C

Capacity of water is 4.18 J/g °C.

Hence, the  quantity of heat is required to raise = mass × Capacity  × raise in temperature

= 450 × 4.18 × (85 - 35) joule

= 94050 joule.

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Answer:

The electric potential energy of the system is 7.87x10⁻²⁰ J.

Explanation:

The electric potential energy is given by:

[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]

Where:

q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C

r is the distance = 3x10⁻⁹ m

K: is the electrostatic constant = 9x10⁹ Nm²/C²

[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]

Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.

I hope it helps you!

The electric potential energy of the system should be 7.87x10⁻²⁰ J.

SInce We know that

fdr = kq1q2/r

Here

q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C

r should be the distance = 3x10⁻⁹ m

K should be the electrostatic constant = 9x10⁹ Nm²/C²

Now electric potential energy should be

= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) /  3x10⁻⁹ m

=  7.87x10⁻²⁰ J.

hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.

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Answer:

The average force exerted on the superball by the sidewalk is 0.00122 N.

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)

final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)

time of motion, t = 1800 s

The average force exerted on the superball by the sidewalk is given by;

[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]

Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.

x = 1/2 at²

where x = length of runway, a = acceleration, and t = time.

600 m = 1/2 (12 m/s²) t²

t² = (1200 m) / (12 m/s²)

t² = 100 s²

t = 10 s

The complete question is;

A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?

A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?

B) What is the power dissipated in his body?

C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?

Answer:

A) I = 1.379 A

B) P = 19016.41 W

C) r = 15990000 Ω

Explanation:

A) We are given;

Internal resistance of the power supply; r = 1600 Ω

Body resistance between hands; R = 10kΩ = 10000 Ω

Power supply voltage; E =16 kV = 16000 V

Formula for the current through the person's body with internal resistance is given by;

I = E/(R + r)

Thus;

I = 16000/(10000 + 1600)

I = 1.379 A

B) Formula for power dissipated is;

P = I²R

P = 1.379² × 10000

P = 19016.41 W

C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A

Thus, from I = E/(R + r) and making r the subject, we have;

r = (E/I) - R

r = (16000/0.001) - 10000

r = 15990000 Ω

Explanation:

Let east = E, and, west = opposite to east = - E.

Here, displacement:

=> 2m east + 4m west + 1m east

=> 2E + 4(-E) + 1E

=> 2E - 4E + 1E

=> - 1E

=> 1(-E)

=> 1m west

And, distance,

=> 2m + 4m + 1m = 7m

The distance of a person is 7 m and the displacement of the person is 1m west.

To find the distance and displacement, the given values are,

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.

Displacement:

Distance:

Let us consider East = E and west = opposite to east = - E.

Calculating the displacement:

= 2m east + 4m west + 1m east

= 2E + 4(-E) + 1E

= 2E - 4E + 1E

= - 1E

= 1(-E)

= 1m west.

The displacement is 1m west.

Now calculating the distance,

= 2m + 4m + 1m

= 7m

The distance is 7m.

Thus, the displacement and the distance is found as 1 m west and 7m.

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]

Explanation:

From the question we are told that

The weight of the mountain climber is m = 555 N

Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as  

       [tex]T_a*  cos 65 -555 + T_b * cos(85) =  0[/tex]

Here  [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side

 So

    [tex]0.423T_a   + 0.0871T_b  =  555[/tex]

=>   [tex] 0.0871T_b  =  555 - 0.423T_a[/tex]

=>   [tex] T_b  =  \frac{555 - 0.423T_a}{0.0871}[/tex]

Generally from the diagram , the total amount of force acting on the rope along the horizontal  axis at equilibrium is mathematically represented as

      [tex]T_a*  sin 65 - T_b * sin(85) =  0[/tex]

=>     [tex] 0.9063T_a - 0.9962T_b =  0[/tex]

=>     [tex] 0.9063T_a =   0.9962T_b [/tex]

=>     [tex] 0.9063T_a =   0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]

=>     [tex] 0.9063T_a =   [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]

=>    [tex] 0.0789T_a =   [552.891 - 0.421T_a[/tex]

=>    [tex] 0.4999T_a =   552.891 [/tex]

=>      [tex] T_a = 1106 \ N [/tex]

Answer: 11400 m

Explanation:

Given:

t = 3.8 x 10^-5 s

v = 3 x 10^8 m/s

d = ?

Formula:

d = vt

  = (3.8 x 10^-5 s)(3 x 10^8 m/s)

  = 11400 m

hope this helps :)

The uniform movement allows to find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km

Kinematics studies the motion of objects looking for relationships between position, velocity and acceleration, in the special case that the acceleration is zero is called uniform motion and is described by the expression

                [tex]v = \frac{d}{t}[/tex]

                d = v t

Where v is the velocity, d the displacement and t the time.

Radar waves are electromagnetic waves with constant velocity

            v = 3 10⁸ m/s

They indicate that the time of the waves to go to the plane and return is 3.8 10⁻⁵ s, therefore if the speed is constant, the time to reach the plane is half of the total time.

          t = [tex]\frac{t_{total} }{ 2}[/tex]

          t = [tex]\frac{3.8 \ 10^{-5}}{2}[/tex]  

          t = 1.9 10⁻⁵ s

Let's calculate

        d = 3 10⁸ 1.9 10⁻⁵

        d = 5.7 10³

In conclusion with the uniform movement we can find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km

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Answer:

retina.

Explanation:

Answer:

Your parrot, from earth, that weighs 1.33kg is 1.33kg on earth. as far as i'm aware there is only one earth and everything always weighs the same on one planet as it did on that same planet.

Explanation:

Answer:

A ) E/8

Explanation:

If the sphere of radius R  carries charge Q,  then the volumetric charge density is:

ρ₁ = [Q/ (4/3)*π*R³]

Therefore the net charge inside r  ( r < R ) is:

q₁ = ρ * (4/3)*π*r³

And E = K * q₁/r                  K = 9,98 *10⁹ [N*m²/C²]

E = K *  ρ * (4/3)*π*r³/r

E = K *  ρ * (4/3)*π*r²

If now the charge is distributed over a sphere of radius 2R

ρ₂ =  [Q/ (4/3)*π*(2R)³]

ρ₂ =  [Q/ (4/3)*π*8*R³]

Then  ρ₂ < ρ₁    in fact     ρ₂ = (1/8)*ρ₁

The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then

E₂ = E₁/8

The right answer is lyrics  A ) E/8

Answer:

9.05m/s

Explanation:

given data

m1= 2700kg

v1=3.7m/s

m2=0.18kg

v2=9.05m/s

v3=?

We know that the velocity of the rhino will remains unchanged after impact as the mass of the rubber ball is negligible

m1v1+m2v2=m1v1+m2v3

2700*3.7+0.18*9.05=2700*3.7+0.18*v3

9990+1.629=9990+0.18v3

9991.629-9990=0.18v3

1.629=0.18v3

v3=1.629/0.18

v3=9.05m/s

Answer: the long day in neptune would be .18383562 years!

Explanation:also for every day is 16 hours

Answer:

Yes at A the mechanical energy is conserved.

Yes at B the part of mechanical energy is conserved potential energy and  kinetic energy and some is lost as frictional force.

Explanation:

Ratio = Eb/ Ea=  1058.3 J/2940 J= 0.3599

Ratio = Ec/ Eb= 0J/ 1058.3 J= 0

At point A the skater is at rest  or it is the starting point and the whole energy is due to the position of the  skater i.e= mgh = 50 *9.8*6=  2940 J

Since there's no movement there is no Kinetic energy = 0 J

Yes at A the mechanical energy is conserved.

At point B the skater has traveled for some of the distance . It has potential energy and  kinetic energy.

Yes at B the part of mechanical energy is conserved as potential energy and  kinetic energy.

The total Mechanical energy = 1058.3 J

At point B Total Mechanical energy = PE+ KE

1058.3J = 980 J + 78.3 J

1058.3 J = mgh + 1/2mv²

      = 50*2*9.8 + 1/2 *50*(8.85)²

       = 980 J + 78.3 J

As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .

2940 J-1058.3 J= 1881.7

At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME  all are zero.

(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.

(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.

(c) mechanical energy is conserved between a and b.

(d) mechanical energy is not conserved between b and c.

The given parameters;

The ratio of the mechanical energy at B and mechanical energy at A;

[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]

The ratio of the mechanical energy at C and mechanical energy at A;

[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]

Thus, we can conclude that mechanical energy is conserved between a and b.

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]

Thus, we can conclude that mechanical energy is not conserved between b and c.

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Answer:

(a) 0.829 m/s

(b) 3.27 m/s

(c) 0.000153 m²

55.8%

Explanation:

(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)

Q = vA

(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)

v = 0.829 m/s

(b) Use Bernoulli equation.  Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head.  Choose 0 elevation to be at point 1.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)

1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa

v = 3.27 m/s

(c) Flow rate is constant.

Q = vA

(0.001 m³ / 2.00 s) = (3.27 m/s) A

A = 0.000153 m²

Flow rate is proportional to the pressure difference and the radius raised to the fourth power.

Q ∝ ΔP r⁴

Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴

Q₂/Q₁ = (1.120) (0.840)⁴

Q₂/Q₁ = 0.558

The flow decreases to 55.8% of the original value.

Answer:

Explanation:

Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed.  As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.

In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.

The new flow rate = (1.12)*(0.84)^4

=0.5576 or 55.76% of the original flow rate

Answer:

The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

Explanation:

The value of g can be found using the following equation:

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

[tex] ma = \frac{GmM}{r^{2}} [/tex]

[tex] a = \frac{GM}{r^{2}} [/tex]

Where:

a is the acceleration of gravity = g

G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)

M: is the Earth's mass = 5.97x10²⁴ kg

r: is the Earth's radius = 6371 km      

Since we need to find g at 6700 m, the total distance is:

[tex] r_{T} = 6371000 m + 6700 m = 6377700 m [/tex]

Now, the value of g is:

[tex] a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2} [/tex]

Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

I hope it helps you!

Answer:

The two charged objects will exert equal and opposite forces on each other.

Explanation:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.

This force of attraction or repulsion between the two charged objects is always equal and opposite.

Therefore, the two charged objects will exert equal and opposite forces on each other.

Answer:

I chose c because it is the greater slope at point c

Answer:

Average speed

Explanation:

250/50=5

1000/20=5

Answer:

A. Current power may decrease the fish population.

Explanation:

The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.

The environment is made up of living and non-living components that co-exist and interact with one another.

Answer:

A

Explanation:

I took the test good luck :D

Answer:

light reflects when it hits the surface

Explanation:

Youngs double slit is a evidence for wave nature,

The properties refraction are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.

So,

     the answer must be B

Answer: Light does not need a medium to travel.

Explanation: I took the test and got it right :]

given info is... Acceleration(a)=2.6m/s^2

                       final velocity(v)=26.8m/s

                       initial velocity(u)=24.6m/s

need to find.... time(t)=?

[tex]a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\[/tex]

[tex]t=\frac{v-u}{a}[/tex]

[tex]t=\frac{26.8-24.6}{2.6}[/tex]

[tex]t=0.846s[/tex]

Explanation:

It takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.

Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.

Given parameters:

Initial speed of the car: u = 24.6 m/s

Final speed of the car: v = 26.8 m/s.

Acceleration of the car: a = 2.6 m/s²

Time interval: t = ?

change is speed = final speed - initial speed

= 26.8 m/s - 24.6 m/s

= 2.2 m/s

From the definition of acceleration,

acceleration = change is speed / time interval

So, time interval  =  change is speed / acceleration

= 2.2 m/s/2.6 m/s²

= 0.84 second.

Hence, it takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.

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Answer:

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)

Where:

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

[tex]k[/tex] - Spring constant, measured in newtons per meter.

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

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Answer:

apex consumers

Explanation:

they are top

Answer:

The average force exerted on the car during this time is 5,943 N

Explanation:

Given;

mass of the car, m = 849 kg

initial velocity of the car, u = 0

time of motion of the car, t = 5.00 s

final velocity of the car, v = 35 m/s

The average force exerted on the car during this time is given by;

[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]

Therefore, the average force exerted on the car during this time is 5,943 N

Answer:

5943N

Let's say (+x) = eastward

Average horizontal acceleration

ax = vx -v0x/5.00s

= 35.0m/s-0/5.00s

= +7.09m/s

From here we apply the second law of newton

During this period average horizontal force acting on car

Summation x = max = (849kg)(+7.09m/s²)

= 5943N

+5.943x10³N

= 5.94kN east ward.

Answer:speeding up constantly

Explanation:

The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.

A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.

A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.

Given:

The graph shows the speed of a car traveling east over a 12-second period,

As you can see from the graph, at time t = 0 sec the speed is 10 m/s,

At t = 3 sec, the speed = 15.3 m/s

At t = 6 sec, the speed = 20.3 m/s

Thus, speed is increasing constantly.

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Answer:

d, Due to all reasons in a, b and c​

Explanation:

All of the reasons from the choices can lead to errors in measurement. An error is an uncertainty introduced to a scientific process.

Errors can be due to the following reasons;

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force