A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 56.0 km/h in the 18.0 s that it takes to round the bend. The radius of the curve is 160 m. Compute the acceleration at the moment the train speed reaches 56.0 km/h. Assume the train continues to slow down at this time at the same rate.

Answer:

mass of the second ball is 0.379m

Explanation:

Given;

mass of first ball = m

let initial velocity of first ball = u₁

let final velocity of first ball  = v₁ = 0.45u₁

let the mass of the second ball = m₂

initial velocity of the second ball, u₂ = 0

let the final velocity of the second ball = v₂

Apply the principle of conservation of linear momentum;

mu₁ + m₂u₂ = mv₁ + m₂v₂

mu₁  +  0  = 0.45u₁m + m₂v₂

mu₁  = 0.45u₁m + m₂v₂ -------- equation (i)

Velocity for elastic collision in one dimension;

u₁ + v₁ = u₂ + v₂

u₁ + 0.45u₁ = 0 + v₂

1.45u₁ = v₂ (final velocity of the second ball)

Substitute in v₂ into equation (i)

mu₁  = 0.45u₁m + m₂(1.45u₁)

mu₁ = 0.45u₁m + 1.45m₂u₁

mu₁ - 0.45u₁m = 1.45m₂u₁

0.55mu₁ = 1.45m₂u₁

divide both sides by u₁

0.55m = 1.45m₂

m₂ = 0.55m / 1.45

m₂ = 0.379m

Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)

The mass of the second ball is 0.379m and this can be determined by conserving the momentum.

Given :

A ball of mass 'm' makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its original speed.

In order to determine the mass of the second ball, apply conservation of linear momentum.

[tex]\rm m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

where [tex]m_1[/tex] is the mass of the first ball, [tex]m_2[/tex] is the mass of the second ball, [tex]u_1[/tex] is the initial velocity of the first ball, [tex]u_2[/tex] is the initial velocity of the second ball, [tex]\rm v_1[/tex] is the final velocity of the first ball, and [tex]\rm v_2[/tex] is the final velocity of the second ball.

Now, substitute the known terms in the above formula.

[tex]\rm mu_1+0=0.45u_1m+m_2v_2[/tex]

[tex]\rm mu_1=0.45u_1m+m_2v_2[/tex]   ----  (1)

For elastic collision, the velocity is given by:

[tex]\rm v_1+u_1=v_2+u_2[/tex]

[tex]\rm 0.45u_1+u_1=0+v_2[/tex]

[tex]\rm v_2 = 1.45u_1[/tex]

Now, substitute the value of [tex]\rm v_2[/tex] in the equation (1).

[tex]\rm mu_1=0.45u_1m+1.45u_1m_2[/tex]

[tex]\rm 0.55mu_1=1.45m_2u_1[/tex]

[tex]\rm m_2=0.379m[/tex]

So, the mass of the second ball is 0.379m.

For more information, refer to the link given below:

https://brainly.com/question/19689434

Answer:

4.9 m/s

Hope this helps! ;)

Explanation:

Answer:

Both attempt to explain human behavior

Explanation:

Psychology is generally regarded as the science of human behavior. Behaviourism is the psychological theory which holds that behaviour can be fully understood in terms of conditioning, without actually considering thoughts or feelings. The theory holds that psychological disorders can be aptly handled by simply altering the behavioural patterns of the individual. It involves the study of stimulus and responses.

Cognitive psychology attempts to decipher what is going on in people's minds. That is, it looks at the mind as a processor of information. Hence we can define cognitive psychology as the study of the internal mental processes. This according to behaviorists, cannot be studied in measurable terms as in behaviourism (stimulus response approach) even though mental processes are known to influence human behavior significantly.

Hence, both behaviourism and cognitive psychology attempt to study human behavior from different perspectives.

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:

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:

X = R,   therefore the current and the voltage are in phase, that is, the angles are zero

Explanation:

In circuits with capacitors and inductors the phase between current and voltage varies according to frequency , capacitance and induction values, but they go in opposite directions. The first creates a delay in the current and the second an advance.

   If it is an RLC type serial circuit, the impedance is

            X = √[ R² + (XL- Xc)²

with

    XL = wL

    Xc = 1 / wc

let's apply this to mention if XL = Xc, the impedance the circuit is resistive

             X = R

therefore the current and the voltage are in phase, that is, the angles are zero

Answer:

The amount of liters of fuel should have been added is  20093 L

Explanation:

Given that;

a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane

i.e the volume of fuel left in the plane = 7682 L

Required fuel to make  a trip = 22,300 kg of fuel

Also from the question; we are being told that in order for the pilot to determine the volume ; he asked for the density of the fuel and the mechanic said 1.77.

This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L).

Now; we can understand that the density of the fuel was 1.77 pound /litre.

SO , let convert 1.77 pound /litre to kg/Litre;

we all know that

1 pound = 0.4536 kg

1.77 pound/litre  = x kg

If we cross multiply ; we will have:

1.77 pound/litre  × 0.4536 kg = 1 pound × x kg

x kg = (1.77 pound/litre  × 0.4536 kg) /1 pound

x = 0.802872 kg/litre

[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]

where ;

mass =  22,300 kg of fuel

volume = unknown ???

density = 0.802872 kg/litre

making volume the subject of the formula from above; we have:

[tex]\mathbf{volume = \dfrac{mass}{Density}}[/tex]

[tex]\mathbf{volume = 22300 \ kg \ of \ fuel *\dfrac{1 \ litre }{0.802872 \ kg \ of \ fuel}}[/tex]

volume = 27775.28672 litre

volume [tex]\approx[/tex] 27775 L

Let not forget that we are being told as well that the volume of fuel left in the plane = 7682 L

Now;

The amount of liters of fuel should have been added is: =  27775 L - 7682 L

The amount of liters of fuel should have been added is  20093 L

Answer:

Final speed = 2.067 m/s

Explanation:

We are told that the child weighs 26 kg.

Also, that the wagon weighs 5kg.

Thus,initial mass of the child and wagon with ball is;

m_i = 26 + 5 = 31 kg.

Also, we are told that the child now dropped 1.5 kg ball from the wagon. So,

Final mass is;

m_f = 26 + 5 - 1 = 30 kg

Now, from conservation of linear momentum, we know that;

Initial momentum = final momentum

Thus;

m_i * v_i = m_f * v_f

Where v_i is initial velocity and v_f is final velocity.

Making v_f the subject, we have;

v_f = (m_i * v_i)/m_f

We are given that initial velocity v_i = 2 m/s

Plugging in the relevant values, we have;

v_f = (31 * 2)/30

v_f = 2.067 m/s

Answer:

a

  [tex]PE = 2.3 *10^{-16} \ J[/tex]

b

 [tex]T = 1.1 *10^{7} \ K[/tex]

Explanation:

From the question we are told that

      The distance of separation is  [tex]d = 1.00 *10^{-12} \ m[/tex]

Generally the electric potential energy can be mathematically represented as

            [tex]PE = \frac{k * q_1 q_2 }{d}[/tex]

Given that in a nuclei the only charged particle is the proton who charge is

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

Hence

     [tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]

And k is the coulomb constant with values   [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]

      So we have that

       [tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]

      [tex]PE = 2.3 *10^{-16} \ J[/tex]

The relationship between the electrical potential energy and the temperature is mathematically represented as

         [tex]PE = \frac{3}{2} kT[/tex]

Here  k is  the Boltzmann's constant with value  [tex]k = 1.38*10^{-23} JK^{-1}[/tex]

   making T the subject

       [tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]

substituting values

      [tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]

     [tex]T = 1.1 *10^{7} \ K[/tex]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

[tex]C^2=R^2+D^2[/tex]

[tex]C^2=(3D)^2+D^2[/tex]

[tex]C^2=9D^2+D^2[/tex]

[tex]C^2=10D^2[/tex]

[tex]C=\sqrt{10D^2}=3.2D[/tex]

[tex]\frac{C}{D}=3.2[/tex]

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

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

Hydraulic Lift - After change

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

Where:

[tex]P[/tex] - Hydrostatic pressure, measured in pascals.

[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.

[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.

[tex]F[/tex] - Hydrostatic force, measured in newtons.

[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

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

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

[tex]\Delta F = A\cdot \Delta P[/tex]

Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:

[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]

[tex]\Delta F = 2500\,N[/tex]

The additional mass needed for the additional weight is:

[tex]\Delta m = \frac{\Delta F}{g}[/tex]

Where:

[tex]\Delta F[/tex] - Additional weight, measured in newtons.

[tex]\Delta m[/tex] - Additional mass, measured in kilograms.

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

If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:

[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\Delta m = 254.92\,kg[/tex]

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

[tex]\Delta P \propto \rho \cdot \Delta h[/tex]

[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]

Where:

[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.

[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]

[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]

Where:

[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.

[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.

If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:

[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]

[tex]\Delta h_{air} = 158.140\,m[/tex]

The height of the building is 158.140 meters.

158.13m

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ([tex]P_{R}[/tex])

Pressure measured at the ground level =  ([tex]P_{G}[/tex])

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) [tex]P_{G}[/tex] =  ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

[tex]P_{G}[/tex] =  13600 x 0.76 x 10

[tex]P_{G}[/tex] = 103360 Pa

(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g       --------------(**)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

[tex]P_{R}[/tex]  =  13600 x 0.745 x 10

[tex]P_{R}[/tex]  = 101320 Pa

(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = [tex]\frac{2040}{12.9}[/tex]

h = 158.13m

Therefore, the height of the building is 158.13m

Answer:

[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Explanation:

Given that

Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]

slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]

slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]

We know that for double slit the maxima condition is that

[tex]\operatorname{dsin} \theta=m \lambda[/tex]

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

[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]

For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]

[tex]\tan \theta=\frac{y_{m}}{D}[/tex]

[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]

Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]

Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]

Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

[tex]\triangle E = 12.79 J[/tex]

Explanation:

Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]

Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]

Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]

Difference in maximum stored energy between the sprinters and the non-athlethes:

[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

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:

no two electrons in the same atom can have the same set of four quantum numbers

Explanation:

Pauli 's Theory of Exclusion specifies that for all four of its quantum numbers, neither two electrons in the same atom can have similar value.

In a different way, we can say that no more than two electrons can take up the identical orbital, and two electrons must have adversely spin in the identical orbital

Therefore the second option is correct

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:

Explanation:

For resistances in parallel we know that the overall resistance will be smaller in the value than any individual resistance. so, in this case we are concerned about [tex]\frac{1}{R}[/tex] = [tex]\frac{1}{Rs}[/tex] ⁺ [tex]\frac{1}{Rp}[/tex]

where Rs is the series resistor and Rp is the parallel resistor.

so, R = 0.9M, roughly.

So, as the infernal resistance is very high as compared to the resistance conntected in parallel that is 100 Ω is not be used.

The digital multimeter (DMM) can have its internal resistance determined without taking into account the 100 Ω resistor because the internal resistance is very high when compared to that of the parallel connection, and this does not change as the resistance principle remains the same.

Generally, Singles connection of resistors when connected, it topples the resultant parallel connection of resistors.

Therefore

[tex]\frac{1}{R} = \frac{1}{Rs} * \frac{1}{Rp}[/tex]

Hence, because the internal resistance (R)is very high when compared to that of the parallel (Rp), the 100ohms is not put to use.

In conclusion, This could be done for a resistor of any resistance value as the resistance principle remains the same.

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Answer:

l these errors believe that the speed of the system is less than that calculated

Explanation:

When we carry out any measurement in addition to the magnitude, the sources of uncertainty must also be analyzed.

We can have random uncertainties, correspondin

g to momentary errors, for example early warps during medicine, parallax errors, errors in the starting and ending points of the movement; I mean every possible random error. This error is the one that is analyzed and calculated in the statistical equations

There is another source of error, the systematic ones, these are much more complicated, they can be an error in the pendulum length, friction in the pendulum movement mechanism, deformities in the support systems, this errors are not analyzed by the statistic, in general They discover by looking at the results and comparing with the tabulated or real ones.

tith the explanation we see that the errors described are systematic.

In general these errors believe that the speed of the system is less than that calculated

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:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

Answer:

F = 0.078N

Explanation:

In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:

[tex]B=\frac{\mu_oNi}{L}[/tex]         (1)

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

N: turns of the solenoid = 580

i: current in the solenoid = 36A

L: length of the solenoid = 18cm = 0.18m

You replace the values of all parameters in the equation (1):

[tex]B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T[/tex]

Next, you calculate the force exerted on the wire, by using the following formula:

[tex]F=iLBsin\theta[/tex]         (2)

i: current in the wire = 27A

L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m

θ: angle between wire and the direction of B

B: magneitc field in the solenoid = 0.145T

The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.

You replace the values of the parameters in the equation (2):

[tex]F=(27A)(0.02m)(0.145T)sin90\°=0.078N[/tex]

The magnitude of the force on the wire is 0.078N

Answer: The magnitude of the force is 0.079N

Explanation: Please see the attachments below

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 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:

Differences between regular and irregular objects are:

Hope this helps...

Good luck on your assignment..

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:

 I = 89 N s

Explanation:

Momentum is a concept that tells us how much the amount of movement of a system changes, it is described by the expression

      I = ∫ F dt = F_average   t

where F is the force and t is the time

let's calculate

      I = 445  0.2

      I = 89 N s

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