A charging bull elephant with a mass of 5500 kg comes directly toward you with a speed of 4.70 m/s . You toss a 0.160-kg rubber ball at the elephant with a speed of 7.50 m/s(a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball's kinetic energy has increased?

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:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ  = 60/(25×9.8)

μ = 60/245

μ = 0.245

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

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:

1.99*10-4sec

Explanation:

Signal propagation speed=0.82∗2.46∗108m/s

d=2000 m

Tp=20000/0.82∗2.46∗108 sec

ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8

= 1.99* 10^-4seconds

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:

[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]

Explanation:

Hello,

In this case, since pressure is defined as the force applied over a surface:

[tex]P=\frac{F}{A}[/tex]

We can associate the force with the weight of the needle computed by using the acceleration of the gravity:

[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]

And the area of the the tip (circle) in meters:

[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]

Thus, the pressure exerted on the record turns out:

[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]

Which is truly a large value due to the tiny area on which the pressure is exerted.

Best regards.

Answer:

Explanation:

Hope this helped,

Kavitha

Answer:

at an advanced stage of social and cultural development. "a civilized society"

Explanation:

polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)

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:

A. b) Directed towards center

B. [tex]v = 7.854\ m/s[/tex]

C. [tex]a_c = 12.337\ m/s^2[/tex]

D. [tex]w = 1.57\ rad/s[/tex]

Explanation:

The "force" that they feel pressing their backs against the wall is because the reaction to the  centripetal acceleration .

A.

This acceleration has its direction towards the center of the circle. (option b)

B.

Their linear speed can be calculated with the equation:

[tex]v = (\theta/t)*r[/tex]

Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:

[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]

C.

The centripetal acceleration is given by the equation:

[tex]a_c = v^2/r[/tex]

[tex]a_c = 7.854^2/5[/tex]

[tex]a_c = 12.337\ m/s^2[/tex]

D.

Their angular speed is given by the equation:

[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]

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:

a) 3344 N

b) 3344 N

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = 3344 N

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is 3344 N

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

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

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  [tex]v = 11.76 \ m/ s[/tex]

Explanation:

From the question we are told that

   The  total distance traveled is  [tex]d = 1.2 \ m[/tex]

    The mass of the block is  [tex]m_b = 0.3 \ kg[/tex]

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   [tex]PE = KE[/tex]

Where  PE  is the potential energy which is mathematically represented as

      [tex]PE = m * g * h[/tex]

substituting values

     [tex]PE = 3 * 9.8 * 0.60[/tex]

      [tex]PE = 17.64 \ J[/tex]

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          [tex]KE = \frac{1}{2} * m v^2[/tex]

So

      [tex]\frac{1}{2} * m* v ^2 = PE[/tex]

substituting values  

  =>    [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]

=>       [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]

=>    [tex]v = 11.76 \ m/ s[/tex]

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: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:

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)

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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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:

T = 2.06h

Explanation:

In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:

[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex]         (1)

T: time for a complete orbit = ?

r: radius of the orbit

G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2

Mm: mass of the moon = 7.34*10^22 kg

The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:

[tex]r=R_m+160km\\\\[/tex]

Rm: radius of the moon = 1737.1 km

[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]

Then, you replace all values of the parameters in the equation (1):

[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]

In hours you obtain:

[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]

The time that the Apollo takes to complete an orbit around the moon is 2.06h

Answer:

Here, "v" is the velocity of electron and "V" is the potential.

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: 479. 425 N

Explanation: the calculation of a body in an elevator obeys Newton law. When it is accelerating upward, the scale reading is greater than the true weight of the person.

It is given by N= m(g+a)

When it is accelerating downward, the scale reading is less than the true weight.

It so given by N = m(g-a)

The answer to the above questions is in the attached photo

Answer:

the scale will read 476.414 N

Explanation:

Weight = 635 N

mass = (weight) ÷ (acceleration due to gravity 9.81 m/^2)

mass m = 635 ÷ 9.81 = 64.729 kg

initial acceleration of the elevator a = 2.45 m/s^2

the force produced by the acceleration of the elevator downwards = ma

your body inertia force try to counteract this force, by a force equal and opposite to the direction of this force, leading to an apparent weight loss

apparent weight = weight - ma

apparent weight = 635 - (64.729 x 2.45)

apparent weight =  635 - 158.586  = 476.414 N

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:

Explanation:

A )

electric field E = V / d where V is potential difference between plates separated by distance d .

putting the given values

E = 370 / .040  V / m

= 9250 V / m

B )

Force on charged particle of charge q in electric field E

F = q E

F = 2.4 x 10⁻⁹ x 9250

= 22200 x 10⁻⁹

= 222 x 10⁻⁷ N .

C ) since field is uniform , force will be constant

work done by electric field putting up this force

= force x displacement

= 222 x 10⁻⁷  x 40 x 10⁻³

= 888 x 10⁻⁹ J

D )

change in potential energy

= q ( V₁ - V₂ )

= 2.40 X 10⁻⁹ x 370

= 888 x 10⁻⁹ J .

(a) The magnitude of electric field in the region between the plates is 9,250 V/m.

(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].

(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].

The given parameters;

(a) The magnitude of electric field in the region between the plates is calculated as;

[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]

(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;

F = Eq

F = 9,250 x 2.4 x 10⁻⁹

F = 2.22 x 10⁻⁵ N

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;

[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]

(d) the change of the potential energy is calculated as;

[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]

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Answer:

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.

Friction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.

Pressure exists as force over an area: P=F/A

Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

let's divide P₁ by P₂

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that, we can say:

P₁=(4/0.9025)P₂=4.4P₂ or

P₂=(0.9025/4)P₁=0.23P₁

Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,

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Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N