Consider an evacuated rigid bottle of volume V that is surrounded by the atmosphere at pressure P0 and temperature T0. A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle. The air trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle. The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere. Determine the net heat transfer through the wall of the bottle during this filling process in terms of the properties of the system and the surrounding atmosphere.

Answer:

water contracts as it freezes at 0°C

Answer:

weeve

Explanation:

Answer:

The friend on moon is richer.

Explanation:

The value of acceleration due to gravity changes from planet to planet. So the weight of 1 Newton of gold carries different mass on different places. So we need to calculate the mass of gold that both persons have.

FRIEND ON MOON:

W₁ = m₁g₁

where,

W₁ = Weight of Gold won by friend on moon = 1 N

m₁ = mass of gold won by friend on moon = ?

g₁ = acceleration due to gravity on moon = 1.625 m/s²

Therefore,

1 N = m₁(1.625 m/s²)

m₁ = 0.62 kg

ON EARTH:

W₂ = m₂g₂

where,

W₂ = Weight of Gold won by me on Earth = 1 N

m₂ = mass of gold won by me on Earth = ?

g₂ = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

1 N = m₁(9.8 m/s²)

m₁ = 0.1 kg

Since, the friend on moon has greater mass of gold than me.

Therefore, the friend on moon is richer.

Answer:

When the magnet is leaving the loop

Explanation:

According to Lenz's law the direction of an induced current in a conductor will oppose the effect which produces it. As the current is induced in the wire loop and force is exerted on the magnet, the force between the magnet and the loop will be attractive when the magnet is leaving the loop because it's is the one that produces the effect which create the current.

Answer:

Yes we can induce current in the coil by moving the magnet in and out of the coil steadily.

Explanation:

A current can be induced there using the magnetic field and the coil of wire. Moving the bar magnet around the coil can induce a current and this is called electromagnetic induction.

The generation of an electromotive force  across an electrical conductor in a fluctuating magnetic field is known as electromagnetic or magnetic induction.

Induction was first observed in 1831 by Michael Faraday, and James Clerk Maxwell mathematically named it Faraday's law of induction. The induced field's direction is described by Lenz's law.

Electrical equipment like electric motors and generators as well as parts like inductors and transformers have all found uses for electromagnetic induction.

Here, moving the bar magnet around the coil generates the electronic movement followed by a generation of electric current.

Find more on electromagnetic induction :

https://brainly.com/question/13369951

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The question is incomplete, however, the correct question is attached

in the image format:

Answer:

B. 171 N

Explanation:

The equation of the forces along the

Horizontal direction:

[tex]T_{2} cos62^{0} = T_{1} cos20^{0}[/tex]...... 1

Verticalb direction:

[tex]T_{1} sin20^{0} = T_{2} sin62^{0}[/tex] = W . . . 2

Where W = 180 N is the weight of the box.

From equation (1),

[tex]= T_{1} =T_{2} \frac{cos62^{0}}{ cos20^{0}}[/tex]

Substituting into equation (2),

[tex](T_{2} \frac{cos62^{0}}{ cos20^{0}})[/tex][tex]sin20^{0} = T_{2} sin62^{0}[/tex]

= [tex]T = \frac{W}{cos62x^{0} tan20x^{0}+sin62x^{0} }[/tex]

=117 N

Thus, the correct answer is option B. 117 N

Answer:

Their translational kinetic energies are the same

Explanation:

The translational kinetic energy of an object is given by the formula:

[tex]KE = 0.5 mv^2[/tex]

Where m = the mass of the object and

v = the linear speed of the object

From the question, it is stated that wheel A has the same mass as wheel B, that is [tex]m_A = m_B[/tex]

Linear speed is also a function of the distance covered. Since both wheels cover the same distance within the same interval, we can conclude that [tex]v_A = v_B[/tex]

Both wheels A and B have equal speed and mass, this means that their translational kinetic energy is the same.

Note that translational kinetic energy is not a function of the radius

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]      (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex]      (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex]       (3)

The length of the incline is 46.54 m

Answer:

the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]

Explanation:

GIven that:

Bonnie and Clyde are sliding a 325 kg bank safe across the floor to their getaway car.

So ,let assume they are sliding the  bank safe on an horizontal direction

Clyde  →  Δ(bank safe)  → Bonnie

Also; from the above representation; let not forget that the friction force [tex]F_{friction}[/tex] is acting in the opposite direction ←

where;

[tex]F_{friction}[/tex]  = [tex]\mu_k mg[/tex]

The safe slides with a constant speed

If Clyde pushes from behind with 377 N of force while Bonnie pulls forward on a rope with 353 N of force.

Thus; since the safe slides with a constant speed if the two conditions are met; then the net force acting on the slide will be equal to zero.

SO;

[tex]F_{net} = F_{Cylde} + F_{Bonnie} - F_{frition}[/tex]

[tex]F_{net} = F_{Cylde} + F_{Bonnie} - \mu_k \ mg[/tex]

Since the net force acting on the slide will be equal to zero.

Then; [tex]F_{net} =0[/tex]

Also; let [tex]F_{Cylde} = F_c[/tex] and [tex]F_{Bonnie} = F_B[/tex]

Then;

[tex]0 = F_c + F_B - \mu_k \ mg[/tex]

[tex]\mu_k \ mg= F_c + F_B[/tex]

[tex]\mu_k = \dfrac{F_c + F_B}{\ mg}[/tex]

where;

[tex]F_c = 377 \ N \\ \\ F_B = 353 \ N \\ \\ mass (m) = 325 \ kg[/tex]

Then;

[tex]\mu_k = \dfrac{377 + 353}{325*9.81}[/tex]

[tex]\mu_k = \dfrac{730}{3188.25}[/tex]

[tex]\mathbf{\mu_k =0.2290}[/tex]

Thus; the safe's coefficient of kinetic friction on the bank floor is [tex]\mathbf{\mu_k =0.2290}[/tex]

Answer:

a) v₁ = 3.92 m / s , b)     ΔP =  = 9.0 10⁴ Pa, c)  t = 0.0297 s  

Explanation:

This is a fluid mechanics exercise

a) let's use the continuity equation

let's use index 1 for the hose and index 2 for the nozzle

        A₁ v₁ = A₂v₂

in area of ​​a circle is

       A = π r² = π d² / 4

we substitute in the continuity equation

        π d₁² / 4 v₁ = π d₂² / 4 v₂

        d₁² v₁ = d₂² v₂

the speed of the water in the hose is v1

       v₁ = v₂ d₂² / d₁²

       v₁ = 14 (1.8 / 3.4)²

        v₁ = 3.92 m / s

b) they ask us for the pressure difference, for this we use Bernoulli's equation

       P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2

as the hose is horizontal y₁ = y₂

       P₁ - P₂ = ½ ρ (v₂² - v₁²)

      ΔP = ½ 1000 (14² - 3.92²)

       ΔP = 90316.8 Pa = 9.0 10⁴ Pa

c) how long does a tub take to flat

the continuity equation is equal to the system flow

        Q = A₁v₁

        Q = V t

where V is the volume, let's equalize the equations

         V t = A₁ v₁

         t = A₁ v₁ / V

A₁ = π d₁² / 4

let's reduce it to SI units

         V = 120 l (1 m³ / 1000 l) = 0.120 m³

          d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m

let's substitute and calculate

         t = π d₁²/4   v1 / V

         t = π (3.4 10⁻²)²/4 3.92 / 0.120

         t = 0.0297 s

Answer:

Force

Explanation:

Force is simply known as pull or push of an object

Answer:

N = 13.65 rpm

Explanation:

given data

radius = 11 m

centripetal acceleration =  2.3 times that due to gravity

to find out

how many revolutions per minute

solution

we know here centripetal accel = 2.3 × g

ω²r  = 2.3 × 9.8

ω² × 11  = 2.3 × 9.8  

solve it we get

ω² = 2.0490

ω = 1.43 rad/s

and

ω = [tex]\frac{2\pi N}{60}[/tex]

1.43 = [tex]\frac{2\pi N}{60}[/tex]  

solve it we get

N = 13.65 rpm

Answer:

xtotal = 90km

displacement = 18km N

Explanation:

To find the total distance traveled by the car, you first calculate the distance traveled by the car when it travels to north. You use the following formula:

[tex]x=vt[/tex]    (1)

x: distance

v: speed of the car = 30 m/s

t: time = one half hour

In order to calculate the distance you convert the time from hours to seconds:

[tex]t=0.5\ h*\frac{3600s}{1\ h}=1800s[/tex]

Then, you replace the values of t and v in the equation (1):

[tex]x=(30m/s)(1800s)=54000m[/tex]     (2)

Next, you calculate the distance traveled by the car when it travels to south:

[tex]x'=v't'\\\\v'=40\frac{m}{s}\\\\t'=15\ min[/tex]

You convert the time from minutes to seconds:

[tex]t'=15\ min*\frac{60s}{1min}=900s[/tex]

[tex]x'=(40m/s)(900s)=36000m[/tex]

Finally, you sum both distances x and x':

[tex]x_{total}=x+x'=54000m+36000m=90000m=90km[/tex]

The total distance traveled by the car is 90km

The total displacement is the final distance of the car respect to the starting point of the motion. This is calculated by subtracting x' to x:

[tex]d=x-x'=54000m-36000m=18000m=18km[/tex]

The total displacement of the car is 18km to the north from its starting point of motion.

Answer:

Changes of state. The kinetic theory of matter can be used to explain how solids, liquids and gases are interchangeable as a result of increase or decrease in heat energy. ... If it is cooled the motion of the particles decreases as they lose energy.13 Nov 2000

Explanation:

Answer:

The answer is electromagnetic

Answer:

electromagnetic

Explanation:

edge 2021

Answer:

Cost per year = $131.4

Explanation:

We are given;

Power rating of computer with monitor;P = 300 W = 0.3 KW

Cost of power per KWh = 15 cents = $0.15

Time used per day by the computer with monitor = 8 hours

Thus; amount of power consumed per 8 hours each day = 0.3 × 8 = 2.4 KWh per day

Thus, for 365 days in a year, total amount amount of power = 2.4 × 365 = 876 KWh

Now, since cost of power per KWh is $0.15, then cost for 365 days would be;

876 × 0.15 = $131.4

Answer:

a) v = 11.24 m / s ,    θ = 17.76º   b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

     p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

      [tex]p_{x f}[/tex]= (m₁ + m₂) vx

       p₀ₓ =  p_{x f}

       m₁ v₁ = (m₁ + m₂) vₓ

     vₓ = m1 / (m₁ + m₂) v₁

let's calculate

       vₓ = - 900 / (900 + 1200) 25

       vₓ = - 10.7 m / s

Axis y

initial moment

      [tex]p_{oy}[/tex]= m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

     p_{fy}= (m₁ + m₂) [tex]v_{y}[/tex]

     p_{oy} = p_{fy}

     m₂ v₂ = (m₁ + m₂) v_{y}

     v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

    [tex]v_{y}[/tex] = 1200 / (900+ 1200) 6

    [tex]v_{y}[/tex]  = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

      v = √ (vₓ² + v_{y}²)

      v = RA (10.7²2 + 3,428²2)

      v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

      tan θ = [tex]v_{y}[/tex] / Vₓ

      θ = tan-1 v_{y} / vₓ)

      θ = tan -1 (3,428 / 10.7)

       θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

      Kf = 1/2 (m1 + m2) v2

      K₀ = ½ m₁ v₁² + ½ m₂ v₂²

      Kf = ½ (900 + 1200) 11.24 2

      Kf = 1.3265 105 J

      K₀ = ½ 900 25²  + ½ 1200 6²

      K₀ = 2,8125 10⁵ + 2,16 10₅4

        K₀ = 3.0285 105J

the wasted energy is

        Kf / K₀ = 1.3265 105 / 3.0285 105

        Kf / K₀ = 0.4380

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

P.E = 0.068 J = 68 mJ

Answer:

s = 15.84m

Explanation:

In order to calculate the distance traveled by the bucket, you first use the formula for the torque exerted on the pulley by the weight of the bucket:

[tex]\tau=I\alpha[/tex]        (1)

I: moment of inertia of the pulley

α: angular acceleration of the pulley

You can calculate the angular acceleration by taking into account that the torque is also:

[tex]\tau=Wr[/tex]     (2)

W: weight of the bucket = Mg = (1.9kg)(9,8m/s^2) = 18.62N

r: radius of the pulley = 0.561m

[tex]\tau=(18.62N)(0.561m)=10.44Nm[/tex]

The moment of inertia is given by:

[tex]I=\frac{1}{2}M_pr^2[/tex]     (3)

Mp: mass of the pulley = 8kg

[tex]I=\frac{1}{2}(8kg)(0.561m)^2=1.25kg.m^2[/tex]

You solve the equation (1) for α and replace the values of the moment of inertia and the torque to obtain the angular acceleration:

[tex]\alpha=\frac{\tau}{I}=\frac{10.44Nm}{1.25kgm^2}=8.35\frac{rad}{s^2}[/tex]

Next, you use the following formula to find the angular displacement:

[tex]\theta=\frac{1}{2}\alpha t^2[/tex]

[tex]\theta=\frac{1}{2}(8.35rad/s^2)(2.6s)^2=28.24rad[/tex]

Finally, you calculate the arc length traveled by the pulley, this arc length is equal to the vertical distance traveled by the bucket:

[tex]s=r\theta =(0.561m)(28.24rad)=15.84m[/tex]

The distance traveled by the bucket is 15.84m

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  [tex]E_o = 6.995 *10^{-6} \ V/m[/tex]

Explanation:

From the question we are told that

     The height of the satellite is  [tex]r = 35000 \ km = 3.5*10^{7} \ m[/tex]

      The power output of the satellite is [tex]P = 1 \ KW = 1000 \ W[/tex]

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     [tex]I = \frac{P}{4 \pi r^2}[/tex]

substituting values

      [tex]I = \frac{1000}{4 * 3.142 (3.5*10^{7})^2}[/tex]

      [tex]I = 6.495*10^{-14} \ W/m^2[/tex]

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          [tex]I = c * \epsilon_o * E_o^2[/tex]

Where [tex]E_o[/tex] is the amplitude of the electric field vector of the satellite broadcast so

         [tex]E_o = \sqrt{\frac{2 * I}{c * \epsilon _o} }[/tex]

substituting values

          [tex]E_o = \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }[/tex]

           [tex]E_o = 6.995 *10^{-6} \ V/m[/tex]

Answer:

ΔL = 1.653 km

Explanation:

The linear expansion of any object due to change in temperature is given by the following formula:

ΔL = αLΔT

where,

ΔL = Change in length or expansion of steel pipe line = ?

α = coefficient of linear expansion of steel = 12 x 10⁻⁶ /°C

L = Original Length of the steel pipe = 1300 km

ΔT = Change in temperature = 35°C - (- 71°C) = 35°C + 71°C = 106°C

Therefore,

ΔL = (12 x 10⁻⁶ /°C)(1300 km)(106°C)

ΔL = 1.653 km

Answer:

(b) electrons will not be ejected

Explanation:

Determine the number of photons ejected by 10 W red laser pointer.

The wavelength (λ) of red light is  700 nm = 700 x 10⁻⁹ m

Energy of a photon is given as;

[tex]E = \frac{hc}{\lambda}[/tex]

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

c is speed of light, = 3 x 10⁸ m/s

[tex]E = \frac{6.626*10^{-34} *3*10^8}{700 X 10^{-9}} \\\\E = 2.8397 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 10 W red laser pointer

10 W = 10 J/s

[tex]Number \ of \ photons = 10(\frac{ J}{s}) * \frac{1}{2.8397*10^{-19}} (\frac{photon}{J} ) = 3.522 *10^{19} \ photons/s[/tex]

Determine the number of photons ejected by 5 W red green pointer

The wavelength (λ) of green light is  500 nm = 500 x 10⁻⁹ m

[tex]E = \frac{hc}{\lambda} = \frac{6.626*10^{-34} *3*10^8}{500*10^{-9}} = 3.9756 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 5 W green laser pointer

5 W = 5 J/s

[tex]Number \ of \ photons = \frac{5J}{s} *\frac{photon}{3.9756*10^{-19}J} = 1.258 *10^{19} \ Photons/s[/tex]

The number of photons emitted by 10 W red laser pointer is greater than the number of photons emitted by 5 W green laser pointer.

Thus, 5 W green laser pointer will not be able to eject electron from the same metal.

The correct option is "(b) electrons will not be ejected"

Answer:

4.1 seconds

Explanation:

The height of the football is given by the equation:

[tex]H = -16t^2 + V*t + S[/tex]

Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):

[tex]0 = -16t^2 + 64*t + 4[/tex]

[tex]4t^2 - 16t - 1 = 0[/tex]

Using Bhaskara's formula, we have:

[tex]\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272[/tex]

[tex]t_1 = (-b + \sqrt{\Delta})/2a[/tex]

[tex]t_1 = (16 + 16.49)/8 = 4.06\ seconds[/tex]

[tex]t_2 = (-b - \sqrt{\Delta})/2a[/tex]

[tex]t_2 = (16 - 16.49)/8 = -0.06\ seconds[/tex]

A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.

The amount of time the football spent in air before it hits the ground is 4.1 s.

The given parameters;

To find:

Using the vertical model equation given as;

[tex]H = -16t^2 + Vt + S\\\\[/tex]

the final height when the ball hits the ground, H = 0

[tex]0 = -16t^2 + 64t + 4\\\\16t^2 - 64t - 4 = 0\\\\divide \ through \ by\ 4\\\\4t^2 - 16t - 1= 0\\\\solve \ the \ quadratic \ equation \ using \ the \ formula \ method;\\\\\\a = 4, \ b = -16, \ c = - 1\\\\t = \frac{-b \ \ + /- \ \ \ \sqrt{b^2 - 4ac} }{2a} \\\\[/tex]

[tex]t = \frac{-(-16) \ \ + /- \ \ \ \sqrt{(-16^2 )- 4(4\times -1)} }{2\times 4}\\\\t = \frac{16 \ \ + /- \ \ \sqrt{272} }{8} \\\\t = \frac{16 \ \ +/- \ \ 16.49}{8} \\\\t = \frac{16 - 16.49}{8} \ \ \ \ or \ \ \ \frac{16 + 16.49}{8} \\\\t = -0.61 \ s \ \ or \ \ \ 4.06 \ s\\\\t\approx 4.1 \ s[/tex]

Thus, the amount of time the football spent in air before it hits the ground is 4.1 s.

Learn more here: https://brainly.com/question/2018532

Answer:

It is the force that opposes motion between two surfaces touching each other. ( OR ) The force between two surfaces that are sliding or trying to slide across each other.

Explanation:

Answer:

a constant force that acts on objects that rub together

Explanation:

a constant force that acts on objects that rub together

Answer:

A charged atom or particle is called an ion :)

Answer:

The difference is [tex]\Delta B = 1.39 \ T[/tex]

Explanation:

From the question we are told that  

    The inner radius  is  [tex]r_i = 0.700 \ m[/tex]

        The outer radius  is  [tex]r_o = 1.20 \ m[/tex]

        The number of turns is  [tex]N = 900 \ turns[/tex]

        The current on each wire is [tex]I = 13.0 kA = 13*10^{3} \ A[/tex]

Generally magnetic field of a toroid along the outer radius is mathematically evaluated as

        [tex]B_o = \frac{\mu_o * N * I}{2 \pi r_o}[/tex]

Where  [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o= 4\pi * 10^{-7} N/A^2[/tex]

substituting values

            [tex]B_o = \frac{ 4\pi * 10^{-7} * 13*10^{3} * 900}{ 2 * 3.142 * 1.20}[/tex]

           [tex]B_o = 1.95 \ T[/tex]

Generally magnetic field of a toroid along the inner radius is mathematically evaluated as

           [tex]B_i = \frac{\mu_o * N * I}{2 \pi r_i}[/tex]

substituting values

           [tex]B_i = \frac{ 4\pi * 10^{-7} * 900 * 13*10^{3}}{2 *3.142 *0.700}[/tex]

         [tex]B_i = 3.34 \ T[/tex]

The difference in  magnitudes of the magnetic fields of the toroid along the inner and outer radii is mathematically evaluated as

      [tex]\Delta B = B_i - B_o[/tex]

      [tex]\Delta B = 3.34 -1.95[/tex]

      [tex]\Delta B = 1.39 \ T[/tex]

Answer:

a) vo = 14m/s

b) v = 14m/s

c) t = 2.85s

d) t = 0.829s

e) v =  22.12 m/s

Explanation:

a) To find the initial velocity of the ball yo use the following formula:

[tex]h_{max}=\frac{v_o^2}{2g}[/tex]         (1)

hmax:  maximum height reached by the ball but measured from the point at which the ball is thrown = 25.0m - 15.0m = 10.0m

vo: initial velocity of the ball = ?

g: gravitational acceleration = 9.8m/s^2

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\sqrt{2gh_{max}}}=\sqrt{2(9.8m/s^2)(10.0m)}=14\frac{m}{s}[/tex]

The initial velocity of the ball is 14m/s

b) To find the velocity of the ball when it is at the same position as the initial point where it was thrown, you can use the following formula:

[tex]v^2=2gh_{max}\\\\v=\sqrt{2gh_{max}}[/tex]        

as you can notice, v = vo = 14m/s

The velocity of the ball is 14 m/s

c) The flight time of the ball is given by twice the time the ball takes to reach the maximum height. You use the following formula:

[tex]t=2\frac{v_o}{g}=2\frac{14m/s}{9.8m/s^2}=2.85s[/tex]             (3)

The time is 2.85s

d) To find the time the ball takes to arrive to the ground after the ball passes the same height at which is was thrown, you can use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]          (4)

y: 0 m (ball just after it impact the ground)

yo: initial position = 15.0 m

vo: in)itial velocity of the ball = 14m/s    

t: time

You replace the values of the parameters in the equation (4) and obtain a quadratic formula:

[tex]0=15.0-14t-\frac{1}{2}(9.8)t^2\\\\[/tex]

You use the quadratic formula to find the roots t:

[tex]t_{1,2}=\frac{-(-14)\pm\sqrt{(-14)^2-4(4.9)(15)}}{2(-4.9)}\\\\t_{1,2}=\frac{14\pm22.13}{-9.8}\\\\t_1=0.829s\\\\t_2=-2.19s[/tex]

you choose the positive values because is has physical meaning

The time the ball takes to arrive to the ground is 0.829s

e) The final velocity is:

[tex]v=v_o+gt[/tex]

[tex]v=14m/s+(9.8m/s^2)(0.829s)=22.12\frac{m}{s}[/tex]

The final velocity is 22.14 m/s

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

[tex]F_2-F_1-F_f=Ma[/tex]          (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

[tex]a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}[/tex]

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero

Answer:

The electric force is  [tex]F = 11.9 *10^{-9} \ N[/tex]

Explanation:

From the question we are told that

    The  Bohr radius at ground state is  [tex]a_o = 0.529 A = 0.529 ^10^{-10} \ m[/tex]

    The values of the distance between the proton and an electron  [tex]z = 2.63a_o[/tex]

The electric force is mathematically represented as

     [tex]F = \frac{k * n * p }{r^2}[/tex]

Where n and p are charges on a single electron and on a single proton which is mathematically represented as

      [tex]n = p = 1.60 * 10^{-19} \ C[/tex]

    and  k is the coulomb's  constant with a value

           [tex]k =9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.[/tex]

substituting values

       [tex]F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}[/tex]

         [tex]F = 11.9 *10^{-9} \ N[/tex]

Answer:

a) [tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex], b) Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex], c) The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].

Explanation:

a) The total energy of the object is equal to the sum of potential and kinetic energies. That is:

[tex]E = K + U[/tex]

Where:

[tex]K[/tex] - Kinetic energy, dimensionless.

[tex]U[/tex] - Potential energy, dimensionless.

After replacing each term, the total energy of the object at any point in its motion is:

[tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex]

b) The amplitude of the motion occurs when total energy is equal to potential energy, that is, when objects reaches maximum or minimum position with respect to position of equilibrium. That is:

[tex]E = U[/tex]

[tex]E = \frac{1}{2} \cdot k \cdot A^{2}[/tex]

Amplitude is finally cleared:

[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]

Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex].

c) The maximum speed of the motion when total energy is equal to kinetic energy. That is to say:

[tex]E = K[/tex]

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

Maximum speed is now cleared:

[tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex]

The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].