a uniform rod of 30cm is pivoted at its center.a 40N weight is hung 5cm from left.from where 50N weight be hung to maintain equilibrium?

Answers

Answer 1

Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium


Related Questions

Which has more mass electron or ion?

Answers

an ion has more mass than an electron

A flashlight is held at the edge of a swimming pool at a height h = 1.6 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 1.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.

Required:
What is the horizontal distance from the edge of the pool to the bottom of the pool where the light strikes? Write your answer in meters.

Answers

one of the answers that i found was   5.83 m i did some more research and it showed the same answer again. good luck with it. hope i was able to help you.

The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?

Answers

Answer:

F= 175.5N

Explanation:

Given:

Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance

torque of 38N⋅m

angle of 120∘

Torque(τ): 38Nm

position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m

Angle: 120°

To find the Force, the torque equation will be required which is expressed below

τ = Frsinθ

We need to solve for F, if we rearrange the equation, we have the expression below

F= τ/rsinθ

Note: the torque is maximum when the angle is 90 degrees

But θ= 180-120=60

F= 38/0.25( sin(60) )

F= 175.5N

Two parallel plates have charges of equal magnitude but opposite sign. What change could be made to increase the strength of the electric field between the plates

Answers

Answer:

The electric field strength between the plates can be increased by decreasing the length of each side of the plates.

Explanation:

The electric field strength is given by;

[tex]E = \frac{V}{d}[/tex]

where;

V is the electric potential of the two opposite charges

d is the distance between the two parallel plates

[tex]E =\frac{V}{d} = \frac{\sigma}{\epsilon _o} \\\\(\sigma = \frac{Q}{A} )\\\\E = \frac{Q}{A\epsilon_o} \\\\E = \frac{Q}{L^2\epsilon_o}[/tex]

Where;

ε₀ is permittivity of free space

L is the length of each side of the plates

From the equation above, the electric field strength can be increased by decreasing the length of each side of the plates.

Therefore, decreasing the length of each side of the plates, could be made to increase the strength of the electric field between the plates

A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev

Answers

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

     

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]

A solenoid 26.0 cm long and with a cross-sectional area of 0.580 cm^2 contains 490 turns of wire and carries a current of 90.0 A.
Calculate:
(a) the magnetic field in the solenoid;
(b) the energy density in the magnetic field if the solenoid is filled with air;
(c) the total energy contained in the coil’s magnetic field (assume the field is uniform);
(d) the inductance of the solenoid.

Answers

Answer:

A.21.3T

B.1.8x 10^6J/m^3

C.0.27x10^2J

D.6.6x10^-3H

Explanation:

Pls see attached file

Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.

Answers

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

select the example that best describes a renewable resource.
A.after a shuttle launch, you can smell the jet fuel for hours.
B.solar panels generate electricity that keeps the satellites running.
C.tractor trailers are large trucks that run on diesel fuel.
D. we use our barbeque every night; it cooks with propane.

Answers

Answer:

B.solar panels generate electricity that keeps the satellites running.

Explanation:

Solar panels are a renewable resource because they take energy from the sun.

An air conditioner connected to a 103 V rms AC line is equivalent to a 20 resistance and a 1.68 inductive reactance in series. a) What is the impedance of the air conditioner

Answers

Answer:

20.07ohms

Explanation:

Impedance is defined as the opposition to the flow of current through the elements of the circuit.

Impedance for R-L AC circuit is expressed as Z = √R²+XL²

R is the resistance

XL is the inductive reactance.

Given resistance of the air condition = 20 ohms

Inductive reactance XL = 1.68 ohms

Z = √20²+1.68²

Z = √400+2.8224

Z = √402.8224

Z = 20.07 ohms

Hence the impedance of the air conditioner is 20.07ohms

A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb

Answers

Answer:

931.00ft-lb

Explanation:

Pls see attached file

The work done in moving the object from x = 1 ft to x = 18 ft is 935  ft-lb.

What is work?

Work is the product of the displacement's magnitude and the component of force acting in that direction. It is a scalar quantity having only magnitude and Si unit of work is Joule.

Given that force = 6x - 2 pounds.

So, work done in moving the object from x = 1 ft to x = 18 ft is = [tex]\int\limits^{18}_1 {(6x-2)} \, dx[/tex]

= [ 3x² - 2x]¹⁸₁

= 3(18² - 1² ) - 2(18-1) ft-lb

= 935  ft-lb.

Hence, the work done is  935  ft-lb.

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A sound wave of frequency 162 Hz has an intensity of 3.41 μW/m2. What is the amplitude of the air oscillations caused by this wave? (Take the speed of sound to be 343 m/s, and the density of air to be 1.21 kg/m3.)

Answers

Answer:

I believe it is 91

Explanation:

A 900 kg roller coaster car starts from rest at point A. rolls down the track, goes
around a loop (points B and C) and then flies off the inclined part of the track (point D),
Figure 2.
The dimensions are: H =80 m.
r= 15m, h=10m and theta =9.30°

Calculate the
(a) gravitational potential energy at point A.

(b) velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J.

c) distance of the car land (in the horizontal direction) from point D if given the
velocity at point D is 37.06 m/s

I​

Answers

Answer:

gravitational potential energy at point A.

A) The gravitational potential energy at point A is; 705600 J

B) The velocity at point C, if the work done to move the roller coaster from point B to C is 264870 J is; v = 31.295 m/s

A) Formula for gravitational potential energy is;

PE = mgh

At point A;

mass; m = 900 kg

height; h = 80 m

Thus;

PE = 900 × 9.8 × 80

PE = 705600 J

B) Kinetic energy of the roller coaster at point C is given as;

KE = PE - W

We are given Workdone; W = 264870 J

Thus;

KE = 705600 - 264870

KE = 440730 J

Thus, velocity at point C is gotten from the formula of kinetic energy;

KE = ½mv²

v = √(2KE/m)

v = √(2 × 440730/900)

v = 31.295 m/s

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At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s

Answers

Answer:

5.81 X 10^3 Ns

Explanation:

Given that

F = At² and F at t = 1.25 s is 781.25 N ?

A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²

d(Impulse) = Fdt

Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s

Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s

Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns

5.81 X 10^3 N.s

A 1000-turn toroid has a central radius of 4.2 cm and is carrying a current of 1.7 A. The magnitude of the magnetic field along the central radius is

Answers

Answer:

0.0081T

Explanation:

The magnetic field B in the toroid is proportional to the applied current I and the number of turns N per unit length L of the toroid. i.e

B ∝ I [tex]\frac{N}{L}[/tex]

B = μ₀ I [tex]\frac{N}{L}[/tex]                   ----------------(i)

Where;

μ₀ = constant of proportionality called the magnetic constant = 4π x 10⁻⁷N/A²

Since the radius (r = 4.2cm = 0.042m) of the toroid is given, the length L is the circumference of the toroid given by

L = 2π r

L = 2π (0.042)

L = 0.084π

The number of turns N = 1000

The current in the toroid = 1.7A

Substitute these values into equation (i) to get the magnetic field as follows;

B = 4π x 10⁻⁷ x  1.7 x  [tex]\frac{1000}{0.084\pi }[/tex]        [cancel out the πs and solve]

B = 0.0081T

The magnetic field along the central radius is 0.0081T

A commercial aircraft is flying westbound east of the Sierra Nevada Mountains in California. The pilot observes billow clouds near the same altitude as the aircraft to the south, and immediately turns on the "fasten seat belt" sign. Explain why the aircraft experiences an abrupt loss of 500 meters of altitude a short time later.

Answers

Answer:

Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.

Explanation:

Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.

Resistance and Resistivity: The length of a certain wire is doubled while its radius is kept constant. What is the new resistance of this wire?

Answers

Answer:

Explanation:

The formula for calculating the resistance of a material in terms of its resistivity is expressed as [tex]R = \rho L/A[/tex] where;

R is the resistance of the material

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the area = πr² with r being the radius

[tex]R = \rho L/\pi r^{2}[/tex]

If the length of a certain wire is doubled while its radius is kept constant, then the new length of the wire L₁ = 2L

The new resistance of the wire R₁ will be expressed as [tex]R_1 = \frac{\rho L_1}{A_1}[/tex]

since the radius is constant, the area will also be the same i.e A = A₁ and the resistivity also will be constant. The new resistance will become

[tex]R_1 = \frac{\rho(2L)}{A}[/tex]

[tex]R_1 = \frac{2\rho L}{\pi r^2}[/tex]

Taking the ratio of both resistances, we will have;

[tex]\frac{R_1}{R} = \frac{2\rho L/\pi r^2}{\rho L/ \pi r^2} \\\\\frac{R_1}{R} = \frac{2\rho L}{\pi r^2} * \frac{\pi r^2}{ \rho L} \\\\\frac{R_1}{R} = \frac{2}{1}\\\\R_1 = 2R[/tex]

This shoes that the new resistance of the wire will be twice that of the original wire

The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.

Answers

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The spring scale [tex]F_2[/tex] reads  [tex]F_2 = 2.4225 \ N[/tex]

Explanation:

From the question we are told that

      The first force is  [tex]F_1 = 10.5 \ N[/tex]

      The acceleration by which the cart moves to the right is  [tex]a = 2.50 \ m/s^2[/tex]

      The mass of the cart is  m  = 3.231  kg

       

Generally the net force on the cart is  

       [tex]F_{net} = F_1 - F_2[/tex]

This net force is mathematically represented as

      [tex]F_{net} = m * a[/tex]

So  

        [tex]m* a = 10 - F_2[/tex]

        [tex]F_2 = 10.5 - 2.5 (3.231)[/tex]

        [tex]F_2 = 2.4225 \ N[/tex]

 

1. The frequency of a wave defines
O A. the minimum height of a wave.
O B. the maximum height of a wave.
O C. how fast the wave is moving in cycles per second.
D. the height of the wave at a given point.

Answers

Answer:

The answer is C

Explanation:

Frequency, in physics, the number of waves that pass a fixed point in unit time; also, the number of cycles or vibrations undergone during one unit of time by a body in periodic motion. ... See also angular velocity; simple harmonic motion.

Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.

Answers

Answer:

Explanation:

Using E= hc/wavelength

6.63 x10^-34 x3x10^8/ 236nm

19.86*10^-26/236*10^-9

=0.08*10^-35Joules

If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B

Answers

Answer:

0.9m/s²

Explanation:

See attached files

collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity

Answers

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

Suppose a particle moves back and forth along a straight line with velocity v(t), measured in feet per second, and acceleration a(t). What is the meaning of ^120∫60 |v(t)| dt?

Answers

Answer:

The meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

Explanation:

We are told that the particle moves back and forth along a straight line with velocity v(t).

Now, velocity is the rate of change of distance with time. Thus, the integral of velocity of a particle with respect to time will simply be the distance covered by the particle.

Thus, the meaning of the integral (120, 60)∫ |v(t)| dt is simply the distance covered by the particle from time t = 60 seconds to time t = 120 seconds

A 18.0 kg electric motor is mounted on four vertical springs, each having a spring constant of 24.0 N/cm. Find the period with which the motor vibrates vertically.

Answers

Answer:

Explanation:

Total mass m = 18 kg .

Spring are parallel to each other so total spring constant

= 4 x 24 = 96 N/cm = 9600 N/m

Time period of vibration

[tex]T=2\pi\sqrt{\frac{m}{k} }[/tex]

Putting the given  values

[tex]T=2\pi\sqrt{\frac{18}{9600} }[/tex]

= .27 s .

A string passing over a pulley has a 3.85-kg mass hanging from one end and a 2.60-kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.5 cm and mass 0.79 kg .
A. If the bearings of the pulley were frictionless, what would be the acceleration of the two masses?
B. In fact, it is found that if the heavier mass is given a downward speed of 0.20 m/s , it comes to rest in 6.4 s . What is the average frictional torque acting on the pulley?

Answers

Answer:

Explanation:

Let the acceleration be a of the system

T₁ and T₂ be the tension in the string attached with 3.85 and 2.6 kg of mass

for motion of 3.85 kg , applying newton's law

3.85g - T₁ = 3.85 a

for motion of 2.6 kg

T₂ - 2.6g = 2.6 a

T₂ - T₁ + 1.25 g = 6.45 a

T₁ - T₂ = 1.25 g -  6.45 a

for motion of pulley

(T₁ - T₂ ) x R = I x α where R is radius of pulley , I is its moment of inertia and α is angular acceleration

(T₁ - T₂ ) x R = 1 /2  m R² x a / R

(T₁ - T₂ )  =   m  x a / 2 = .79 x a / 2 = . 395 a

1.25 g -  6.45 a = .395 a

1.25 g = 6.845 a

a = 1.79 m /s²

B )

When heavier mass is given speed of .2 m /s , it comes to rest in 6.4 s

Average deceleration = .2 / 6.4 = .03125 m /s²

Total deceleration created by frictional torque = 1.79 + .03125

= 1.82125 m /s²

If R be the average frictional torque  acting on the pulley

angular deceleration of pulley = a / R

= 1.82125 / .045

= 40.47 rad /s²

Now  R = I x 40.47 , I is moment of inertia of pulley

= 1 /2 x .79 x .045² x 40.47

= .0323 N.m

Torque created = .0323 Nm

The acceleration of the two masses hanging from ends of the pulley is 31 m/s².

The average frictional torque acting on the pulley is 0.55 Nm.

The given parameters;

mass of the pulley, = M = 0.79 kgfirst mass, m₁ = 3.85 kgsecond mass, m₂ = 2.6 kgradius, R = 4.5 cm = 0.045 m

The acceleration of the two masses is determined by taking net torque acting on the pulley;

[tex]\tau _{net} = I \alpha[/tex]

[tex]T_1R - T_2R = I \alpha\\\\[/tex]

where;

T is the tension on both stings suspending the masses = mgI is the moment of inertia of the pulley [tex]= \frac{MR^2}{2}[/tex]α is the angular acceleration

[tex]R(T_1 - T_2) = (\frac{MR^2}{2} )(\frac{a}{R} )\\\\T_1 - T_2 = (\frac{MR^2}{2} )(\frac{a}{R} ) \times \frac{1}{R} \\\\T_1 - T_2 = \frac{M}{2} \times a\\\\a = \frac{2}{M} (T_1 - T_2)[/tex]

Substitute the given parameters, to solve for the acceleration of the masses;

[tex]a = \frac{2}{M} (m_1g - m_2 g)\\\\a = \frac{2g}{M} (m_1 - m_2)\\\\a = \frac{2 \times 9.8}{0.79} (3.85 - 2.6)\\\\a = 31 \ m/s^2[/tex]

The average frictional torque acting on the pulley when the heavier mass speeds down by 0.2 m/s and stop by 6.4 s.

[tex]a = \frac{v}{t} = \frac{0.2}{6.4} = 0.031 \ m/s^2 \\\\ a_t = 31 m/s^2+ 0.031 m/s^2 = 31.031 m/s^2 \\\\\tau = I \alpha\\\\\tau = (\frac{MR^2}{2} )(\frac{a_t}{R} )\\\\\tau = (\frac{0.79 \times 0.045^2 }{2} ) (\frac{31.031}{0.045} )\\\\\tau = 0.55 \ Nm[/tex]

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A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time

Answers

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

[tex]\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s[/tex]

The induced emf in the shorter coil is calculated as;

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

[tex]E = NA\frac{\delta B}{\delta t}[/tex]

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

The induced emf in the coil at the center of the longer solenoid is [tex]1.725\times10^{-4}V[/tex]

Induced EMF:

The induced emf is produced in a coil when the magnetic flux through the coil is changing. It opposes the change of magnetic flux. Mathematically it is represented as the negative rate of change of magnetic flux at follows:

[tex]E=-\frac{\delta\phi}{\delta t}[/tex]

where E is the induced emf,

[tex]\phi[/tex] is the magnetic flux through the coil.

The changing current varies the magnetic flux through the coil at the center of the long solenoid, which is given by:

[tex]\phi = \frac{\mu_oNIA}{L}[/tex]

so;

[tex]\frac{\delta\phi}{\delta t}=\frac{\mu_oNA}{L} \frac{\delta I}{\delta t}[/tex]

where N is the number of turns of longer solenoid, A is the cross sectional area, I is the current and L is the length of the coil.

[tex]\frac{\delta\phi}{\delta t}=\frac{4\pi \times10^{-7} \times600 \times \pi \times(1.25\times10^{-2})^2}{25\times10^{-2}} \frac{5}{60}\\\\\frac{\delta\phi}{\delta t}=1.23\times10^{-7}Wb/s[/tex]

The emf produced in the coil at the center of the solenoid which has 14 turns will be:

[tex]E=N\frac{\delta \phi}{\delta t}\\\\E=14\times1.23\times10^{-7}V\\\\E=1.725\times10^{-4}V[/tex]

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A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the switch is open and the capacitor is uncharged. What is the voltage across the resistor and the capacitor at the moment the switch is closed

Answers

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

The index of refraction of a sugar solution in water is about 1.5, while the index of refraction of air is about 1. What is the critical angle for the total internal reflection of light traveling in a sugar solution surrounded by air

Answers

Answer:

The critical angle is  [tex]i = 41.84 ^o[/tex]

Explanation:

From the  question we are told that

    The index of refraction of the sugar solution is  [tex]n_s = 1.5[/tex]

   The  index of refraction of air is  [tex]n_a = 1[/tex]

Generally from Snell's  law

      [tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]

Note that the angle of incidence in this case is equal to the critical angle

Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]

So  

      [tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]

      [tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]

      [tex]i = 41.84 ^o[/tex]

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?

Answers

Answer:

a

  [tex]F =326.7 \ N[/tex]

b

  [tex]M = 6[/tex]

Explanation:

From the question we are told that

          The mass of the rock is  [tex]m_r = 200 \ kg[/tex]

          The  length of the small object from the rock is  [tex]d = 2 \ m[/tex]

          The  length of the small object from the branch [tex]l = 12 \ m[/tex]

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      [tex]W = m_r * g[/tex]

substituting values

     [tex]W = 200 * 9.8[/tex]

     [tex]W = 1960 \ N[/tex]

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         [tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]

Here  [tex]\theta[/tex] is very small so  [tex]cos\theta * l = l[/tex]

                               and  [tex]cos\theta * d = d[/tex]

Hence

       [tex]F * l - W * d = 0[/tex]

=>    [tex]F = \frac{W * d}{l}[/tex]

substituting values

        [tex]F = \frac{1960 * 2}{12}[/tex]

       [tex]F =326.7 \ N[/tex]

The  mechanical advantage is mathematically evaluated as

          [tex]M = \frac{W}{F}[/tex]

substituting values

        [tex]M = \frac{1960}{326.7}[/tex]

       [tex]M = 6[/tex]

Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the following function.
h=1/2t2+1/2t
(a) What is the height of the balloon at the end of 40 sec?
(b) What is the average velocity of the balloon between t = 0 and t = 30?
ft/sec
(c) What is the velocity of the balloon at the end of 30 sec?
ft/sec

Answers

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

A)  The height of the balloon at the end of 40 sec is 820 feet.

B) The average velocity of the balloon is 30 ft/sec.

C) The velocity of the balloon at the end of 30 sec is

Velocity

Given :

h=1/2t²+1/2t

Part A)

The height of the balloon after 40 secs is :

h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

Part B)

The average velocity of the balloon is  :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0 sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

When at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon = 30 ft/sec

The average velocity of the balloon  is  30 ft/sec.

Part C)

The velocity of the balloon at the end of 30 sec is :

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec.

Learn more about "Velocity":

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Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus

Answers

Answer:

2.55m

Explanation:

Using 1/do+1/di= 1/f

di= (1/f-1/do)^-1

( 1/0.0500-1/0.0510)^-1

= 2.55m

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