A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.

Answer:

Doppler Radar gathers information about precipitation by sending out pulses of ___Radio wave___ energy

Answer:

Explanation:

according to resultant of two parallel forces,

Fpivot = Fobject + Fbricks

so that, the net force is zero

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast,  a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

Answer:

Explanation:

The magnetic field produced is expressed using the formula

[tex]B = \frac{\mu_0NI}{L}[/tex]

B is the magnetic field = 0.30T

I is the current produced in the coil = 4.5A

[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]

[tex]N = \frac{BL}{\mu_0I}[/tex]

Substituting the give values to get N;

[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]

The number of turns the solenoid must have is approximately 16,931 turns

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

Answer:

the answer is option A = photoelectric effect

Explanation:

If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.

Answer:

a) the final angular speed is 0.738 rad/s

b) the change in kinetic energy = 0.3 J

Explanation:

the two 1 kg objects have a total mass of 2 x 1 = 2 kg

radius of rotation of the objects = 0.9 m

moment of inertial of the student and the chair = 6 kg-m^2

initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s

final angular speed of rotation of the sitting student and object system ω2 = ?

moment of inertia of the rotating object is

[tex]I = mr^{2}[/tex] = 2 x [tex]0.9^{2}[/tex] = 1.62 kg-m^2

total moment of inertia of sitting student and object system will be  

==> 6 + 1.62 = 7.62 kg-m^2

The initial angular momentum of the sitting student and object system will be calculated from

==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2

if the radius of rotation of the object is reduced to 0.39 m,

new moment of inertia of the rotating object will be

[tex]I = mr^{2}[/tex]  = 2 x [tex]0.39^{2}[/tex] = 0.304 kg-m^2

new total moment of inertia of the sitting student and object system will be

==> 6 + 0.304 = 6.304 kg-m^2

The final momentum of the sitting student and object system will be calculated from

==> Iω2 = 6.304 x ω2 = 6.304ω2

According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system. Therefore,

4.65 = 6.304ω2

ω2 = 4.65/6.30 = 0.738 rad/s

b) Rotational kinetic energy of the system = [tex]\frac{1}{2} Iw^{2}[/tex]

for the initial conditions, kinetic energy is

==>  [tex]\frac{1}{2} Iw1^{2}[/tex] =  [tex]\frac{1}{2}* 7.62*0.61^{2}[/tex] = 1.417 J

for the final conditions, kinetic energy is

==>  [tex]\frac{1}{2} Iw1^{2}[/tex] =  [tex]\frac{1}{2}*6.304*0.738^{2}[/tex] = 1.717 J

change in kinetic energy = final KE - initial KE

==> 1.717 - 1.417 = 0.3 J

Answer:

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]

I is the rotational inertia = 16kgm²

[tex]\theta = angular\ displacement[/tex]

[tex]\theta = 2 rev = 12.56 rad[/tex]

[tex]\alpha \ is \ the\ angular\ acceleration[/tex]

To get the angular acceleration, we will use the formula;

[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]

[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

=  257.23/2.0

= 128.61 Watts

Answer:A

Explanation:

Explanation:

Given that,

Gravitational force = 600 N

Frictional force = 25 N

Pulled by the Force = 250 N

We know that,

The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.

The balance equation is along y-axis

The box will not move in y-axis therefore, the net force in the y-axis will be zero.

Hence, The net force in the y-direction will be zero.

Answer:

  I = 0.2934 I₀

Explanation:

The expression that governs the transmission of polarization is

         I = I₀ cos² θ

Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of

        I₁ = I₀ / 2

this is the light that enters the second polarizer

        I = I₁ cos² θ  

we substitute

        I = I₀ / 2 cos² 40

        I = I₀ 0.2934

        I = 0.2934 I₀

Answer:

The additional force is  [tex]F_3 = 120 \ N[/tex]

Explanation:

From the question we are told that

      The horizontal force in one direction is  [tex]F_i = 480 \ N[/tex]

       The horizontal force in the opposite direction is  [tex]F_f = -600 \ N[/tex]

The negative sign shows that it is acting in the opposite direction

Generally for the box to remain stationary the net force on it must be equal to zero  that is  

        [tex]F_1 + F_2 +F_3 = 0[/tex]

Where [tex]F_3[/tex] is the additional force required  

    So

          [tex]F_3 = -F_1 - F_2[/tex]

substituting values

         [tex]F_3 = -480 - [-600][/tex]

        [tex]F_3 = -480 + 600[/tex]

       [tex]F_3 = 120 \ N[/tex]

Answer:

The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion

Explanation:

I hope this will help you:)

Answer:

The work done by both Joel and Jerry is equal to 0 J.

Explanation:

The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,

W = F.d

where,

W = Work Done on the Body

F = Force Applied on the Body

d = displacement covered by the body

In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,

W = F(0)

W = 0 J (For both Joel and Jerry)

Answer:

a. the bottom medium

Explanation:

it has the least index of refraction and hence most rarer.

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

[tex]B=\frac{\mu_oI}{2R}[/tex]        (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex]         (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

Answer:

The product of the object's weight and the horizontal distance between the two positions.

Explanation:

Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.

Answer:

The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Explanation:

Given;

mass of the bigger child, M = 30 kg

mass of the smaller child, m = 20 kg

distance between the two children, d = 3 m

This information can be represented diagrammatically;

                                    3m

         |<------------------------------------------------>|

----------------------------------------------------------------------------

         ↓             x            Δ            3-x           ↓

         20kg                                                 30kg

x is the distance from the pivot point that the small child will sit in order to maintain the balance

Take moment about the pivot;

Clockwise moment = anticlockwise moment

30(3-x) = 20x

90 -30x = 20x

90 = 20x + 30x

90 = 50x

x = 90 / 50

x = 1.8 m

Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

The distance from the pivot point which the small child must sit in order to maintain the balance is 1.8 meters.

Given the following data:

To determine what distance from the pivot point is the small child sitting in order to maintain the balance, we would take moment about a pivot:

Note: The distance of the child from the pivot is equal to [tex]3-n[/tex]

For moment:

Clockwise moment = anticlockwise moment

[tex]30(3-n) = 20n\\\\90-30n=20n\\\\90=20+30n\\\\90=50n\\\\n=\frac{90}{50}[/tex]

n = 1.8 meters

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Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

Explanation:

We have

A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.

The frequency of a pendulum is equal to the no of oscillation per unit time. so,

[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]

Tim period is reciprocal of frequency. So,

[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]

The time period of a pendulum is given by :

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

l is length of pendulum

[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]

So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.

Answer:

The number of turns of the solenoid is 3536 turns

Explanation:

Given;

magnetic field of the solenoid, B = 0.1 T

current in the solenoid, I = 1.8 A

length of the solenoid, L = 8cm = 0.08m

The magnetic field near the center of the solenoid is given by;

B = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length

I is the current in the coil

The number of turns per length is calculated as;

n = B / μ₀I

n = (0.1 ) / (4π x 10⁻⁷ x 1.8)

n = 44203.95 turns/m

The number of turns is calculated as;

N = nL

N = (44203.95)(0.08)

N = 3536 turns

Therefore, the number of turns of the solenoid is 3536 turns

Answer:

vr = 142.90 m/s

the magnitude of its relative velocity is 142.90 m/s

Explanation:

Given;

A plane has an airspeed of 142 m/s (eastward)

vi = 142 m/s

16.0 m/s wind is blowing southward at the same time as the plane is flying

vb = 16.0m/s

Writing the relative velocity vector, we have;

Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.

v = 142i - 16j

The magnitude of the velocity is;

vr = √(vi^2 + vb^2)

vr = √(142^2 + 16^2)

vr = √(20420)

vr = 142.8985654231 m/s

vr = 142.90 m/s

the magnitude of its relative velocity is 142.90 m/s

Answer:

Explanation:

Hope this helped,

Kavitha

Answer:

The magnitude of the magnetic field will act in a direction towards me.

Explanation:

When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.

Answer:

the electric field is pointing horizontal direction and in south direction

Explanation:

In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.

Answer:

Explanation:

We shall apply Bernoulli's equation of flow of liquid

1 / 2 ρ v² + ρ gh + P = constant

For calculating velocity in second floor

A₁ V₁ = A₂ V₂

1  x 2.8 = 2 x V₂

V₂ = 1.4 m /s

1 / 2 ρ v₁² + ρ gh₁ + P₁ = 1 / 2 ρ v₂² + ρ gh₂ + P₂

.5 x 10³ x 2.8² + 10³ x 9.8h₁ + 340 x 10³ = .5 x 10³ x 1.4² + 10³ x 9.8 x h₂ + P₂

P₂ = 3.92 x 10³ + 9.8  x 10³ ( h₁ - h₂ ) + 340 x 10³ - .98 x 10³  

= 3.92 x 10³ - 9.8  x 10³ x 4 + 340 x 10³ - .98 x 10³

= 303.74  x 10³ Pa

= 303.74 kPa .

Answer:

(a) P = 103064 Pa = 103.064 KPa

(b) P = 102476 Pa = 102.476 KPa

Explanation:

(a)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)

Pg = 1764 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1764 Pa

P = 103064 Pa = 103.064 KPa

(b)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)

Pg = 1176 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1176 Pa

P = 102476 Pa = 102.476 KPa

The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.

Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.

In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.

The absolute pressure in the bulb can be calculated using the following formula:

P = P₀ + ρgh

where:

P is the absolute pressure in the bulb,

P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),

ρ is the density of the sauce (1200 kg/m³),

g is the acceleration due to gravity (9.8 m/s²), and

h is the height of the sauce in the tube.

(a) When h = 0.15 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)

P ≈ 1.013 x 10⁵ Pa + 1764 Pa

P ≈ 1.031 x 10⁵ Pa

(b) When h = 0.10 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)

P ≈ 1.013 x 10⁵ Pa + 1176 Pa

P ≈ 1.025 x 10⁵ Pa

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