An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T, respectively. The particle passes out of the electric field, but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm.
Part A. What is the particle's charge-to-mass ratio?
Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron

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:

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Explanation:

An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]

[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]

Where:

[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.

[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.

[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.

If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:

[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]

[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]

[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

[tex]x = \frac{n_{R}}{n_{E}}[/tex]

[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]

[tex]x = 8.5222 \times 10^{-11}[/tex]

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer:

force on larger piston = [tex]\frac{fA}{a}[/tex]

Explanation:

we label the pistons as piston A and piston B

small piston A:

area = a

force = f

large piston B:

area = A

force  = ?

Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.

we know that pressure = force ÷ area

pressure of piston A = [tex]\frac{f}{a}[/tex]

pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]

obeying Pascal's law, the system pressures must be equal. Therefore

[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]

force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.

Answer:

Force

Explanation:

Force is simply known as pull or push of an object

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

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:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

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

#SPJ6

Answer:

F₂ = 925.92 N

Explanation:

In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,

σ₁ = σ₂

F₁/A₁ = F₂/A₂

F₂ = F₁ A₂/A₁

where,

F₂ = Initial force that must be applied to narrow arm = ?

F₁ = Load on Wider Arm to be raised = 12000 N

A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²

A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²

Therefore,

F₂ = (12000 N)(25π cm²)/(324π cm²)

F₂ = 925.92 N

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

Answer:

a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d)  traveling wave , e)

Explanation:

A traveling wave is described by the expression

            y = A sin (kx - wt)

where k is the wave vector and w is the angular velocity

let's examine every situation presented

a) a new faster pulse is generated

A faster pulse should have a higher angular velocity

equal speed is related to the period and frequency

            w = 2π f = 2π / T

therefore in this case the period must decrease so that the angular velocity increases

the correct answer is c narrower

b) Generate a pulse, but move your hand more.

Moving the hand increases the amplitude (A) of the pulse

the correct answer is b higher

c) generates a pulse but the force is tightened

Set means that more tension force is applied to the string, so the velicate changes

       v = √ (T /μ)

the correct answer is b faster

d) move your hand back and forth

in this case you would see a pulse series whose sum corresponds to a traveling wave

e) Advance a frame the movie

in this case the wave will be displaced a whole period to the right

the correct answer is b

f) move your hand faster

the waves will have a maximum fast, so they are closer

answer C

g) decrease wave frequency

Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.

the correct answer is b increases its wavelength

Answer:

The answer is electromagnetic

Answer:

electromagnetic

Explanation:

edge 2021

Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264

Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]

The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]

To compare the power of the two cars, first find the Kinetic Energy each one has:

K.E. for Model A

[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]

K.E. for model B

[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]

[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]

Now, determine Power for each model:

Power for model A

[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]

Power for model B

[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]

Comparing power of model B to power of model A:

[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]

[tex]\frac{P_B}{P_A} =[/tex] 8.5264

Comparing power for each model, power for model B is 8.5264 better than model A.

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

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:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer:

More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.

Answer: According to Newtons second Law of motion ;

F = ma (Force  equals  mass multiplied by acceleration.)

The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force

Explanation:

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].

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.

Since there is a lost in kinetic energy, the collision is inelastic collision.  

m'u'+mu = V(m+m')

[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]  

Where

m' = mass of the moving object = 2 kg

m = mass of the stick = 8 kg,

u' = initial velocity of the moving object = 5 m/s

V = common velocity after collision= ?    

u = 0 m/s (at rest).

put the values in the formula,  

[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]

  kinetic energy before collision

[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]  

kinetic energy after collision

[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]  

Lost in kinetic energy = 25-5 = 20 J

Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.

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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.

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Answer:

P = 5.97 × 10^(5) Pa

Explanation:

We are given;

Mass of balloon;m_b = 4.3 kg

Radius;r = 1.54 m

Temperature;T = 289 K

Density;ρ = 1.19 kg/m³

We know that, density = mass/volume

So, mass = Volume x Density

We also know that Force = mg

Thus;

F = mg = Vρg

Where m = mass of balloon(m_b) + mass of helium (m_he)

So,

(m_b + m_he)g = Vρg

g will cancel out to give;

(m_b + m_he) = Vρ - - - eq1

Since a sphere shaped balloon, Volume(V) = (4/3)πr³

V = (4/3)π(1.54)³

V = 15.3 m³

Plugging relevant values into equation 1,we have;

(3 + m_he) = 15.3 × 1.19

m_he = 18.207 - 3

m_he = 15.207 kg = 15207 g

Molecular weight of helium gas is 4 g/mol

Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles

From ideal gas equation, we know that;

P = nRT/V

Where,

P is absolute pressure

n is number of moles

R is the gas constant and has a value lf 8.314 J/mol.k

T is temperature

V is volume

Plugging in the relevant values, we have;

P = (3802 × 8.314 × 289)/15.3

P = 597074.53 Pa

P = 5.97 × 10^(5) Pa

Answer:

the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]

Increase in volume of the glass =  10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]

[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]

[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

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