Two children, Ahmed and Jacques, ride on a merry-go-round. Ahmed is at a greater distance from the axis of rotation than Jacques. Which of the following are true statements?
A. Ahmed has a greater tangential speed than Jacques.
B. Jacques has a greater angular speed than Ahmed.
C. Jacques has a smaller angular speed than Ahmed.
D. Jacques and Ahmed have the same angular speed.

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

[tex]\theta=\frac{1.22 \lambda}{D}[/tex]

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

Answer:

the kinetic energy lost in the collison is a) 30 J

Explanation:

given data

mass of door m1 = 35 kg

width a = 90 cm = 0.9 m

the mass of ball  m2 = 500 g = 0.5 kg

initial speed of ball  u = 20 m/s

final speed of ball  v = 16 m/s

r = 60 cm = 0.6 m

soluion

we will consider here final angular speed of the door = w

so now we use conservation of angular momentum  that is

Li = Lf    ........................1

that is express as

m2 × u × r = I × w + m2 × v × r

put here value and we get  

0.5 × 20 × 0.6 = [tex](m1 \times \frac{a^2}{12})[/tex] × w + 0.5 × 16 × 0.6

solve it we get

w = 0.508 rad/s

so that here

the kinetic energy lost in the collison,

KE = KE initial - KE final     ..................2

put here value

KE = 0.5 × m2 × u² - (0.5 × I × w² + 0.5 × m2 × v²)

KE = 0.5 × (0.5 × 20² - (35 × 0.9² ÷ 12) × 0.508² - 0.5 × 16²) J

KE = 30 J

the kinetic energy lost in the collison is a) 30 J

Answer:

n_glass = 1.404

Explanation:

In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]         (1)

n1: index of refraction of vacuum = 1.00

θ1: angle of the incident light respect to normal of the surface = 38.0°

n2: index of refraction of glass = ?

θ2: angle of the refracted light in the glass respect to normal = 26.0°

You solve the equation (1) for n2 and replace the values of all parameters:

[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]

The index of refraction of the glass is 1.404

Answer is given below

Explanation:

given data

we will consider here

Ping-Pong ball weighs = 3.1 g

diameter =  4.2 cm

solution

Whenever the ball is pushed, the length of the airflow along the outer edge increases and it accelerates. According to Bernoulli's equation. As the speed increases, the pressure decreases, so the pressure at the outer end is reduced. As the pressure at the outer edge is low, the extra air jet pushes it back to the center line.

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

Complete question:

Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.

Answer:

The rate at which the distance between the cars is increasing four hours later is 35 mi/h.

Explanation:

Given;

speed of one car, dx/dt = 28 mi/h South

speed of the second car, dy/dt = 21 mi/h West

The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.

Apply Pythagoras theorem to evaluate this distance;

let the distance between the cars = z

x² + y² = z² -------- equation (1)

Differentiate with respect to time (t)

[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)

Since the speed of the cars is constant, after 4 hours their different distance will be;

x: 28(4) = 112 mi

y: 21(4) = 84 mi

[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]

Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt

[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]

Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h

Explanation:

Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR

Answer:

The tension in the cable is 18371.9 newtons.

Explanation:

Physically speaking, the tension can be calculated with the help of the Second Newton's Law. The upward acceleration means that magnitude of tension must be greater than weight of elevator, whose equation of equilibrium is described below:

[tex]\Sigma F = T - m\cdot g = m \cdot a[/tex]

Where:

[tex]T[/tex] - Tension in the cable, measured in newtons.

[tex]m[/tex] - Mass of the elevator, measured in kilograms.

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

[tex]a[/tex] - Net acceleration of the elevator, measured in meters in square second.

Now, tension is cleared and resultant expression is also simplified:

[tex]T = m \cdot (a + g)[/tex]

If [tex]m = 1700\,kg[/tex], [tex]a = 1\,\frac{m}{s^{2}}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the tension in the cable is:

[tex]T = (1700\,kg)\cdot \left(1\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]T = 18371.9\,N[/tex]

The tension in the cable is 18371.9 newtons.

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.

The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:

While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],

where

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.

However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:

[tex]V = \displaystyle \frac{Q}{C}[/tex].

The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

Answer:

E= 71dB

Explanation:

See attached file for step by step calculation

Answer:

Due to flipping of polarity

Explanation:

During the changing of polarity, the current on the one side is maximum as the polarity change then the current is gradually reducing toget from another end.

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

Answer:39.2J

Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.

Answer:

Final pressure 0.105atm

Explanation:

Let V1 represent the initial volume of dry air at NTP.

under adiabatic condition: no heat is lost or  gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.

Adiabatic expansion:

[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]

273/T2=(5V1/V1)^(1.4−1)

273/T2=5^0.4

Final temperature  T2=143.41 K

Also

P1/P2=(V2/V1)^γ

1/P2=(5V1/V1)^1.4

Final pressure P2=0.105atm

Answer:

Fermat's principle states that the path taken by a ray between two given points is the path that can be traversed in the least time.

Thus this least distance being the same as least time will only apply to reflection alone because in reflection light travels In the same direction so least distance will also mean least time. But for refraction light travels in different mediums at different speeds so least distance and least time paths will definately not be the same

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

[tex]=10^{-5.9}[/tex]

intensity of sound of 5 persons

[tex]I=5\times 10^{-5.9}[/tex]

[tex]dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}[/tex]

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

Answer:

Emin= eV

Explanation:

This minimum amount of energy is the work function and so for an electron to reach the detector the energy must b more than work function and the stopping potential so

Emin=eV

Answer:

P = 1470980 J

Explanation:

We have,

Mass of the hiker is 79 kg

It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.

It is given by :

[tex]P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J[/tex]

So, the potential energy of 1470980 J is associated with a hiker.

Answer:

Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.

Explanation:

Answer:

It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.

. . . is carrying more energy than a wave in the same medium with a lower amplitude.

Answer:

Wmax = 63.65 ≈ 64 lb

Explanation:

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is [tex]\frac{1}{6}[/tex]

So, the equation is

[tex]\frac{1}{6} = 1 - \frac{T_2}{T_1}[/tex]

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is [tex]\frac{1}{3}[/tex]

Therefore the next equation is

[tex]\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}[/tex]

Now solve both the equations

solve equations (1) and (2)

[tex]2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)[/tex]

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

Answer:

T = 480.2N

Explanation:

In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.

The forces on the boxes are:

[tex]T-Mg=0[/tex]      (1)

T: tension of the rope

M: mass of the boxes 0= 49kg

g: gravitational acceleration = 9.8m/s^2

The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.

By using the equation (1) you obtain:

[tex]T=Mg=(49kg)(9.8m/s^2)=480.2N[/tex]

The woman needs to pull the rope at 480.2N

Answer:

2121.2kg/m^3 is the density of the test liquid on the left

Explanation:

See attached file

Answer:

a) q = 4.47 10⁻⁵ C

b)     ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

           U = Q² / 2C

         Q = √ (2UC)

let's reduce the magnitudes to the SI system

   c = 1 nF = 1 10⁻⁹ F

let's calculate

         q = √ (2 1 10⁻⁹-9)

         q = 0.447 10⁻⁴ C

         q = 4.47 10⁻⁵ C

b) for the potential difference we use

             C = Q / ΔV

            ΔV = Q / C

            ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

            ΔV = 4.47 10⁴ V

Answer:

  315,000 ft·lb

Explanation:

At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.

  total work done = work to raise cable + work to raise load

  = (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb

Answer:

7.2N/C

Explanation:

Pls see attached file

Answer:

the angular momentum is 1015.52 kg m²/s

Explanation:

given data

mass of each flywheel, m = 65 kg

radius of flywheel, r = 1.47 m

ω1 = 8.94 rad/s

ω2 = - 3.42 rad/s

to find out

magnitude of the net angular momentum

solution

we get here Moment of inertia that is express as

I = 0.5 m r²    .................1

put here value and we get

I = 0.5 × 65 × 1.47 × 1.47

I = 70.23 kg m²

and

now we get here Angular momentum that is express as

L = I × ω    ...........................2

and Net angular momentum will be

L = 2 × I x ω1 - I × ω2

put here value and we get

L = 2 × 70.23 × 8.94 - 70.23 × 3.42

L = 1015.52 kg m²/s

so

the angular momentum is 1015.52 kg m²/s

The magnitude of the net angular momentum of the system will be "1015.52 kg.m²/s".

According to the question,

Flywheel's mass, m = 65 kg

Flywheel's radius, r = 1.47 m

ω₁ = 8.94 rad/s

ω₂ = 3.42 rad/s

We know,

The moment of inertia (I),

= 0.5 m r²

By substituting the values,

= 0.5 × 65 × 1.47 × 1.47

= 70.23 kg.m²

hence, The angular momentum be:

→ L = I × ω or,

     = 2 × I × ω₁ - l × ω₂

     = 2 × 70.23 × 8.94 - 70.23 × 3.42

     = 1015.52 kg.m²/s

Thus the above answer is correct.

Find out more information about momentum here:

https://brainly.com/question/25121535

Answer:

The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

[tex]F=m\,a[/tex]

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]

Answer:

the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole