What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)

Answer:

Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force,  air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.

Answer:

Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.

Answer:

thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)

Explanation:

i hope this will help you :)

Explanation:

The collision forces are equal and opposite.  Therefore, the magnitudes are equal.

Answer:

15amps

Explanation:

V=IR

I=V/R

I = 120/8

I = 15 amps

Answer:

(c) always equal to the weight of the liquid displaced.

Explanation:

Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.

Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.

Answer:

Explanation:

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]

Where:

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

[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]

[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]

Where:

[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.

[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]

Where:

[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.

[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:

[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]

[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{2} - y_{1} = 56.712\,m[/tex]

Second arrow

[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]

[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{3} - y_{1} = 342.816\,m[/tex]

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

[tex]E = U + K_{x}[/tex]

The expression is now expanded:

[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]

Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.

[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:

[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]

[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]

If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 201.720\,J[/tex]

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

[tex]E = m\cdot g \cdot y_{max}[/tex]

[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]

[tex]E = 201.720\,J[/tex]

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Answer:

spring constant = 31.25N/m

Explanation:

spring constant = force/extension

mass = 250g = 0.25kg

extension = 8cm = 0.08m

force = mg = 0.25 x 10 = 2.5N

spring constant = 2.5/0.08 = 31.25N/m

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  [tex]v = 3.79 *10^{5} \ m/s[/tex]  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]

     The magnetic field is  [tex]B = 0.38 \ T[/tex]

The force due to the electric field is mathematically represented as

      [tex]F_e = E * q[/tex]

and

The force due to the magnetic field is mathematically represented as

    [tex]F_b = q * v * B * sin(\theta )[/tex]

Now given that it is perpendicular ,  [tex]\theta = 90[/tex]

=>   [tex]F_b = q * v * B * sin(90)[/tex]

=>   [tex]F_b = q * v * B[/tex]

Now  given that it is not deflected it means that

        [tex]F_ e = F_b[/tex]

=>    [tex]q * E = q * v * B[/tex]

=>   [tex]v = \frac{E}{B }[/tex]

 substituting values

     [tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]

     [tex]v = 3.79 *10^{5} \ m/s[/tex]

Answer:

electric energy ( power ) = 300000 W

Explanation:

given data

mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s

efficiency = 15 % = 0.15

solution

we know that Efficiency of electric engine is expression  as

Efficiency = Mechanical energy ÷ electric energy  ......................1

and here dam electrical power output is

put here value in equation 1

electric energy ( power ) = Efficiency × Mechanical energy ( power )

electric energy ( power ) = 0.15 × 2000000 J/s

electric energy ( power ) = 300000 W

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

Answer:

in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,

Explanation:

Let's start for ya dream gas

        PV = nRT

Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.

The density of is defined by

        ρ = m / V

As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.

Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.

This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship

             E = h f

Answer:

-0.63 rad/s²

Explanation:

Given that

Initial angular velocity of the turntable, w(i) = 33 1/3 rpm

Final angular velocity of the turntable, w(f) = 0 rpm

Time taken to slow down, t = 5.5 s

The calculation is attached in the photo below

Answer:

The width is  [tex]Z = 0.0424 \ m[/tex]

Explanation:

From the question we are told that

    The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]

    The wavelength of the light is  [tex]\lambda = 721 \ nm[/tex]

      The position of the screen is  [tex]D = 2.83 \ m[/tex]

Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

        [tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]

Where m which is the order of the interference is 1

substituting values

       [tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]

      [tex]\theta = 0.5317 ^o[/tex]

 Now the width of first minimum  of the interference pattern is mathematically evaluated as

       [tex]Y = D sin \theta[/tex]

substituting values

       [tex]Y = 2.283 * sin (0.5317)[/tex]

       [tex]Y = 0.02 12 \ m[/tex]

 Now the width of  the  pattern's central maximum is mathematically evaluated as

        [tex]Z = 2 * Y[/tex]

substituting values

      [tex]Z = 2 * 0.0212[/tex]

     [tex]Z = 0.0424 \ m[/tex]

Answer:

Weight of object on moon is 5N ,as we know. Weight of object on moon is 1/3 the of object on earth,so

let weight of object on earth = X

5N= X/3

X = 3×5 = 15N

Hence the weight of the object on earth will

be 15N

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

(b) Find the angle in radians through which it rotates in this time interval.

Now we convert rad to angle

The angle in radians = 27.9°

Answer:

4.78 second

Explanation:

given data

vertical cliff = 41 m

height = 112 m

solution

we know here time taken to fall vertically from the cliff =  time taken to move horizontally   ..........................1

so we use here vertical component of ball

and that is accelerated motion with initial velocity = 0

so we can solve for it as

height = 0.5 ×  g ×  t²     ........................2

put here value

112 = 0.5 ×  9.8 ×  t²    

solve it we get

t²   = 22.857

t = 4.78 second

ball thrown horizontally from the top of the cliff in 4.78 second

By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.

Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

By  the experiment "By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

To know more about experiment and hydrochloric acid

https://brainly.com/question/13770820

#SPJ3

Answer:

1) Φ=zero

2)  λ = 2π / k

3)   T = 2π / w

4)  v = w / k

Explanation:

The equation of a traveling wave is

     y = A sin (ka - wt + Ф)

Let's answer using this equation the different questions

1) we see that the equation given in the problem the phase is zero

2) wavelength

     k = 2π /λ

      λ = 2π / k

3) The perido

angular velocity is related to frequency

       w = 2π f

frequency and period are related

       f = 1 / T

       w = 2 π / T

        T = 2π / w

4) the wave speed is

         v = λ f

          λ = 2π / k

          f = w / 2π

          v = 2π /k  w /2π

           v = w / k

Answer:

The sun will be located near the Gemini constellation at sunset

Complete question:

A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.

Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain

Answer:

The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.

Explanation:

Given;

magnitude of applied force, F = 1.5 N

Apply Newton's second law of motion;

F = ma

[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;

[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]

Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N

Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.

Answer:

11.7°

Explanation:

See attached file

Answer:

(a)  I = 0.363 kgm^2

(b)  I = 1.95 kgm^2

Explanation:

(a) If you consider the shape of the skater as approximately a cylinder, you use the following formula to calculate the moment of inertia of the skater:

[tex]I_s=\frac{1}{2}MR^2[/tex]         (1)

M: mass of the skater = 60.0 kg

R: radius of the cylinder = 0.110m

[tex]I_s=\frac{1}{2}(60.0kg)(0.110m)^2=0.363kg.m^2[/tex]

The moment of inertia of the skater is 0.363 kgm^2

(b) In the case of the skater with his arms extended, you calculate the moment of inertia of a combine object, given by cylinder and a rod (the arms) that cross the cylinder. You use the following formula for the total moment of inertia:

[tex]I=I_c+I_r\\\\I=\frac{1}{2}M_1R^2+\frac{1}{12}M_2L^2[/tex]       (2)

M1: mass of the cylinder = 74.0 kg

M2: mass of the rod = 3.00kg +3.00kg = 6.00kg

L: length of the rod = 0.750m + 0.750m = 1.50m

R: radius of the cylinder = 0.150

[tex]I=\frac{1}{2}(74.0kg)(0.150m)^2+\frac{1}{12}(6.00kg)(1.50m)^2\\\\I=1.95kg.m^2[/tex]

The moment of inertia of the skater with his arms extended is 1.95 kg.m^2

Answer:

14.79 kgm/s

Explanation:

Data provided in the question

Let us assume the mass of baseball =  m = 0.145 kg

The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s

Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s

Based on the above information, the change in momentum is

[tex]\Delta P = m(v_f -v_i)[/tex]

[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]    

= 14.79 kgm/s

Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s

Given that,

radius = 1.24 m

According to question,

The rope cannot push outwards. It must always have some slight tension or the bucket will fall.

We need to calculate the tension in the rope

At the top the force of gravity is

[tex]F=mg[/tex]

The force needed to move the bucket in a circle is centripetal force.

So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.

The rope have a tension of less than zero.

We need to calculate the velocity of swing bucket

Using centripetal force

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]mg=\dfrac{mv^2}{r}[/tex]

[tex]g=\dfrac{v^2}{r}[/tex]

[tex]v^2=gr[/tex]

[tex]v=\sqrt{gr}[/tex]

Put the value into the formula

[tex]v=\sqrt{9.8\times1.24}[/tex]

[tex]v=3.49\ m/s[/tex]

Hence, The minimum tension in the rope is less than zero .

The bucket swings with the velocity of 3.49 m/s.

Answer:

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

Vs = 0.4 m³

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

r = 0.46 m

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]

Where:

[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.

[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.

[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]

[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]

If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:

[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]

[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]

The speed of the arrow after passing through the target is 30.1 meters per second.

Answer:

fggdfddvdghyhhhhggghh

Answer:

Cost = $ 1.48

Cost = $ 0.005

Cost = $ 0.38

Explanation:

given data

electrical transmission varies = $0.070/kWh to $0.258/kWh

average value = $0.110/kWh

solution

when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be

first we get here energy consumed that is express as

E = Pt    .................1

here E is Energy Consumed and Power Delivered is P and t is time

so power is here 0.04 KW and t = 2 week = 336 hour

so

put value in 1 we get

E = 0.04 × 336

E = 13.44 KWh

so cost will be as

Cost = E × Unit Price    .............2

put here value and we get

Cost = 13.44 × 0.11

Cost = $ 1.48

and

when you making a piece of dark toast in 3.00 min with a 970-W toaster

so energy consumed will be by equation 1 we get

E = Pt

power is = 0.97 KW and time = 3 min = 0.05 hour

put value in equation 1 for energy consume

E = 0.97 × 0.05 h

E = 0.0485 KWh

and we get cost by w\put value in equation 2 that will be

cost =  E × Unit Price

cost = 0.0485 × 0.11

Cost = $ 0.005

and

when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer

from equation 1 we get energy consume

E = Pt

Power Delivered = 5.203 KW and time = 40 min = 0.67 hour

E = 5.203 × 0.67

E = 3.47 KWh

and

cost will by put value in equation 2

Cost = E × Unit Price

Cost = 3.47 × 0.11

Cost = $ 0.38

Answer:

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

[tex]S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\[/tex]

a) An expression for the altitude of the ballast after t seconds is therefore

[tex]S = \frac{1}{2}gt^2\\[/tex]

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

[tex]576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs[/tex]

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft