A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?

Answer:

Explanation:

Find the diagram relating to the question for proper explanation of the question below.

Using the principle of moment

Sum of clockwise moments = Sum of anticlockwise moments

Moment = Force * perpendicular distance

For anti-clockwise moment:

Since the 30 kg moves in the anticlockwise direction according to the diagram

ACW moment = 30 * 1 = 30 kgm

For clockwise moment

If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).

Equating both moments we have;

0.75M = 30

M = 30/0.75

M = 40 kg

The second child's mass is 40 kg

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Complete Question:

A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane, with the cord making a 30° angle with the vertical.

(a) Determine the ball’s speed. (b) If, instead, the ball is revolved so that its

speed is 3.7 m/s, what angle does the cord make with the vertical?

(Check attached image for the diagram.)

Answer:

(a) The ball’s speed, v = 2.06 m/s

(b) The angle the cord makes with the vertical is 50.40⁰

Explanation:

If the ball is revolved in a horizontal plane, it will form a circular trajectory,

the radius of the circle, R = Lsinθ

where;

L is length of the string

The force acting on the ball is given as;

F = mgtanθ

This above is also equal to centripetal force;

[tex]mgTan \theta = \frac{mv^2}{R} \\\\Recall, R = Lsin \theta\\\\mgTan \theta = \frac{mv^2}{Lsin \theta}\\\\v^2 = glTan \theta sin \theta\\\\v = \sqrt{glTan \theta sin \theta} \\\\v = \sqrt{(9.8)(1.5)(Tan30)(sin30)} \\\\v = 2.06 \ m/s[/tex]

(b) when the speed is 3.7 m/s

[tex]v = \sqrt{glTan \theta sin \theta} \ \ \ ;square \ both \ sides\\\\v^2 = glTan \theta sin \theta\\\\v^2 = gl(\frac{sin \theta}{cos \theta}) sin \theta\\\\v^2 = \frac{gl*sin^2 \theta}{cos \theta} \\\\v^2 = \frac{gl*(1- cos^2 \theta)}{cos \theta}\\\\gl*(1- cos^2 \theta) = v^2cos \theta\\\\(9.8*1.5)(1- cos^2 \theta) = (3.7^2)cos \theta\\\\14.7 - 14.7cos^2 \theta = 13.69cos \theta\\\\14.7cos^2 \theta + 13.69cos \theta - 14.7 = 0 \ \ \ ; this \ is \ quadratic \ equation\\\\[/tex]

[tex]Cos\theta = \frac{13.69\sqrt{13.69^2 -(-4*14.7*14.7)} }{14.7} \\\\Cos \theta = 0.6374\\\\\theta = Cos^{-1}(0.6374)\\\\\theta = 50.40 ^o[/tex]

Therefore, the angle the cord makes with the vertical is 50.40⁰

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Answer:

 θ₁ = 0.04º , θ₂ = 0.00118º

Explanation:

The equation that describes the diffraction pattern of a network is

             d sin θ = m λ

where the diffraction order is, in this case they indicate that the order

m = 1

           θ = sin⁻¹ (λ / d)

Trfuvsmod ls inrsd fr ll red s SI units

           d = 12000 line / inc (1 inc / 2.54cm) = 4724 line / cm

the distance between two lines we can look for it with a direct proportions rule

If there are 4724 lines in a centimeter, the distance for two hundred is

            d = 2 lines (1 cm / 4724 line) = 4.2337 10⁻⁴ cm

let's calculate the angles

λ = 300 10-9 m

            θ₁ = sin⁻¹ (300 10-9 / 4,2337 10-4)

            θ₁ = sin⁻¹ (7.08 10-4)

            θ₁ = 0.04º

λ = 5000

          θ₂ = sin-1 (500 10-9 / 4,2337 10-4)

          θ₂ = 0.00118º

Answer:

[tex]\large \boxed{\text{3.3 cm}^{3}}[/tex]

Explanation:

Assume the stone consists of basalt, which has a density of 3.0 g/cm³.

[tex]\rho = \text{10 g}\times\dfrac{\text{1 cm}^{3}}{\text{3.0 g}} = \text{3.3 cm}^{3}\\\\\text{The volume of the stone is $\large \boxed{\textbf{3.3 cm}^{3}}$}[/tex]

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

Answer:

(a)    f = 0.58Hz

(b)    vmax = 0.364m/s

(c)    amax = 1.32m/s^2

(d)    E = 0.1J

(e)    [tex]x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calulated by using the following formula:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]          (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by:

[tex]v_{max}=\omega A=2\pi f A[/tex]     (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by:

[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]

[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is:

[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]

The total energy is 0.1J

(e) The displacement as a function of time is:

[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]

Answer:

2.45 m/s

Explanation:

kinetic energy = 1/2 * m * v^2

then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2

by solving for x, X = 2.45 m/s

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Answer:

E = - dV/dx

Explanation:

Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.

El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.

De los antes expuesto las dos magnitudes están relacionadas

         E = - dV/dx

por lo cual el potenical es el gradiente del potencial eléctrico.

Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial

Answer:

STATIC,  STATIC

KINETIC friction is less than static friction

Explanation:

In this exercise you are asked to complete the sentences with the correct words.

STATIC friction prevents the relative movement of two surfaces in contact.

For moving surfaces the friction is STATIC is greater than the kinetic friction.

For the last two sentences I think they are misspelled, the correct thing is

KINETIC friction is less than static friction

Answer:

-1748*10^N/C

Explanation:

See attached file

Answer:

The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor's magnetic field in 24.95 ms.

Explanation:

The energy stored in the inductor is given as

E₁ = ½LI²

The rate at which energy is stored in the inductor is

(dE₁/dt) = (d/dt) (½LI²)

Since L is a constant

(dE₁/dt) = ½L × 2I (dI/dt) = LI (dI/dt)

(dE₁/dt) = LI (dI/dt)

Rate of Energy dissipated in a resistor = Power = I²R

(dE₂/dt) = I²R

When the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field

(dE₁/dt) = (dE₂/dt)

OK (dI/dt) = I²R

L (dI/dt) = IR

Current in a this kind of series setup of inductor and resistor at any time, t, is given as

I = (V/R) (1 - e⁻ᵏᵗ)

k = (1/time constant) = (R/L)

(dI/dt) = (kV/R) e⁻ᵏᵗ = (RV/RL) e⁻ᵏᵗ = (V/L) e⁻ᵏᵗ

L (dI/dt) = IR

L [(V/L) e⁻ᵏᵗ] = R [(V/R) (1 - e⁻ᵏᵗ)

V e⁻ᵏᵗ = V (1 - e⁻ᵏᵗ)

e⁻ᵏᵗ = 1 - e⁻ᵏᵗ

2 e⁻ᵏᵗ = 1

e⁻ᵏᵗ = (1/2) = 0.5

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = -0.69315

- kt = -0.69315

kt = 0.69315

k = (1/time constant)

Time constant = 36.0 ms = 0.036 s

k = (1/0.036) = 27.78

27.78t = 0.69315

t = (0.69315/27.78) = 0.02495 = 24.95 ms

Hope this Helps!!!

Answer:

Muons reach the earth in great amount due to the relativistic time dilation from an earthly frame of reference.

Explanation:

Muons travel at exceedingly high speed; close to the speed of light. At this speed, relativistic effect starts to take effect. The effect of this is that, when viewed from an earthly reference frame, their short half life of about two-millionth of a second is dilated. The dilated time, due to relativistic effects on time for travelling at speed close to the speed of light, gives the muons an extended relative travel time before their complete decay. So in reality, the muon do not have enough half-life to survive the distance from their point of production high up in the atmosphere to sea level, but relativistic effect due to their near-light speed, dilates their half-life; enough for them to be found in sufficient amount at sea level.  

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

[tex]S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }[/tex]

where

[tex]S_A[/tex] indicates the speed of the ship A

[tex]S_B[/tex] indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

[tex]S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }[/tex]

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

[tex]\theta=90-40=50^o[/tex]

From the triangle OAB

[tex]cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}[/tex]

[tex]1340.57=35^2+25^2-x^2[/tex]

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

Answer:

[tex]r_f=\frac{1}{6}r_i[/tex]

Explanation:

To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.

You use the following expression:

[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex]          (1)

k: Coulomb's constant

qA: charge of A particle

qB: charge of B particle

When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:

[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex]         (2)

To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:

[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]

The new separation of the charges is 1/6 times of the initial separation

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

[tex]i_1+i_3=i_2[/tex]                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

Answer:

208 V

Explanation:

resistor voltage = Vr = 124 V

capacitor voltage Vc = 167 V

the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor

total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]

==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V

Answer:

In collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Explanation:

In a collision two objects, there is a force exerted on both objects that causes an acceleration of both objects. These forces that act on both objects are equal in magnitude and opposite in direction.

Thus, in collision between equal-mass objects, each object experiences the same acceleration, because of equal force exerted on both objects.

Answer:

the acceleration during the collision is: - 5  [tex]\frac{m}{s^2}[/tex]

Explanation:

Using the formula:

[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]

we get:

[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

Answer:

A pie chart is a type of graph in which a circle is divided into sectors that each represents a proportion of the whole

Explanation:

pie charts are a useful way to organize data in order to see the size of components relative to the whole.

Answer:

[tex]v_i = 1.44\frac{m}{s}[/tex]

Explanation:

The computation of the initial velocity of the fly is shown below:-

But before that we need to do the following calculations

For 4.22 seconds

[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]

[tex]= -1.37\frac{m}{s}[/tex]

For uniform acceleration

[tex]\bar v = \frac{v_i +v_f}{2}[/tex]

[tex]= v_i + v_f[/tex]

[tex]= -2.74\frac{m}{s}[/tex]

With initial and final velocities

[tex]= -1.33\frac{m}{s^2}[/tex]

[tex]= \frac{v_i +v_f}{4.22s}[/tex]

[tex]= -v_i + v_f[/tex]

[tex]= -5.61\frac{m}{s}[/tex]

So, the initial velocity is

[tex]v_i = 1.44\frac{m}{s}[/tex]

We simply applied the above steps to reach at the final solution i.e initial velocity

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Answer:

Explanation:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6*10^-19*4*10^3/9.1*10^-31=0.7*10^15 =7*10^14 m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4*10^6[2*2*10^-2/7*10^14]^1/2

=3*10^-2=3 cm

now we find the velocity f the electron strike the plate

v^2-(4*10^6)^2=2*7*10^14*2*10^-2

v^2=16*10^12+28*10^12

v^2=44*10^12

speed after hits =>V=6.6*10^6 m/s

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

[tex]\frac{1}{2}mv^{2}=mgh[/tex]

Let's solve it for v:

[tex]v=\sqrt{2gh}[/tex]

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!