A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

Answer:

The nature of the wave formed is a transverse  progressive wave.

Explanation:

A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.

A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.

The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.

Answer:

i) angle of incidence;i = 29.43°

ii) δm = 38.92°

Explanation:

Prism is equilateral so angle of prism (A) = 60°

Refractive index of glass; n_glass = 1.52

A) Let's assume the incident angle = i and Critical angle = θc

We know that, sin θc = 1/n

Thus;

sin θc = 1/n_glass

θc = sin^(-1) (1/n_glass)

θc = sin^(-1) (1/1.52)

θc = 41.14°

Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.

Thus;

A = r + θc

r = A - θc

So;

r = 60° - 41. 14°

r = 18.86°

From, Snell's law. If we apply it to this question, we will have;

(sin i)/(sin r) = n_glass

Where;

i is angle of incidence and r is angle of reflection.

Let's make i the subject;

i = sin^(-1) (n_glass × sin r)

i = sin^(-1) (1.52 × sin 18.86)

i = sin^(-1) 0.4914

i = 29.43°

B) The formula to calculate minimum deviation would be from;

μ = [sin ((A + δm)/2)]/(sin A/2)

Where;

μ is Refractive index

δm is minimum angle of deviation

A is angle of prism

Now Refractive index is given by a formula; μ = (sin i)/(sin r)

So; μ = (sin 29.43)/(sin 18.86)

μ = 1.52

Thus;

1.52 = [sin ((60 + δm)/2)]/(sin 60/2)

1.52 * sin 30 = sin ((60 + δm)/2)

0.76 = sin ((60 + δm)/2)

sin^(-1) 0.76 = ((60 + δm)/2)

49.46 × 2 = (60 + δm)

98.92 - 60 = δm

δm = 38.92°

Answer:  If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.

Answer:

It will lose electrical potential energy.

Explanation:

A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.

Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.

Answer:

  c.  in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

Explanation:

First law: things keep doing what they are doing, unless force is applied.

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  [tex]\mu mg = \frac{mv^2}{r}[/tex]

    where [tex]\mu[/tex] is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

[tex]h_{max}=\frac{v_o^2}{g}[/tex]   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

[tex]v^2=v_o^2-2gh[/tex]       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]

Then, you solve the previous result for vo:

[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]

The initial speed of the object was 25.45 m/s

Answer:

St Lawrence Sea way

Explanation:

The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.

Answer:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

[tex]\vec{F}=m\vec{a}[/tex]

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]

Both x and y component must be equal in the previous equality, then you have:

[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]

Hence, the vector F3 is:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Answer:

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.

[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]

Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:

[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]

[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]

The angular accelaration measured in radians per square second is:

[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]

[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

[tex]\tau = I \cdot \alpha[/tex]

Where:

[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

Where:

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

[tex]r[/tex] - Radius of the CD, measured in meters.

If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:

[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]

[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]

Now, the net torque exerted on CD is:

[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]

[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Answer:

Explanation:

capacitance of air capacitor

C = ε₀ A /  d

ε₀ is permittivity of medium , A is plate area , d is distance between plate .

C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²

= 170.19 x 10⁻¹⁴ F .

charge on the capacitor when it is charged to  potential of 255 V

= CV , C is capacitance and V is potential

= 170.19 x 10⁻¹⁴  x 255

= 4.34 x 10⁻¹⁰ C .

After it is disconnected from the source , and it is immersed in water , charge on it remains the same .

So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.

b )

When it is immersed in water its capacity increases  k times where k is dielectric constant of water which is 80 .

capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴  F

= 13615.2  x 10⁻¹⁴ F .

= 1.36 x 10⁻¹⁰ F

potential difference = charge / capacitance

= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰

= 3.2 V

c )

Energy of capacitor = 1/2 C V²

Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴

=  55.33 x 10⁻⁹ J

Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²

= .7  x 10⁻⁹ J .

decrease of energy = 54.63 x 10⁻⁹ J .

Answer:

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

Before collision;

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

After collision

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

the answer is 1.5 hope this helps

Answer:

1.5

Explanation:

0.5=σ/(15/0.5)

σ=3/2 or 1.5

Answer:

Explanation:

vertical component of the velocity of arrow

= 26 sin 60 = 22.516 m

height reached by it after 3.99 s

h = ut - 1/2 g t²

= 22.516 x 3.99 - .5 x 9.8 x 3.99²

= 89.83 - 78

11.83 m

Total height of cliff = 1.55 + 11.83

= 13.38 m

c ) maximum height covered s

v² = u² - 2gs

0 = u² - 2gs

s = u² / 2g

= 22.516² / 2 x 9.8

= 25.86

maximum height reached

= 25.86 + 1.55

= 27.41 m

d )

vertical speed after 3.99 s

v = u - gt

= 22.516 - 9.8 x 3.99

= -16.586

Horizontal component will remain unchanged

Horizontal component = 26 cos 60

= 13 m /s

Resultant of two velocities

= √ 13²+ 16.568²

= 21 m /s

Answer:

D

Explanation:

Now the net force is the applied force minus the frictional force; this is expressed mathematically as:

Fnet= Fappplied - Ffrictional

Now the frictional force is given as ;

Coefficient of friction × normal reaction

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction of the human is ;

35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }

Hence the Frictional force =343×0.4 =137.20N

Hence Fnet = 200-137.20 = 62.8N

Fnet = 63N to the nearest whole

The net force on the couch as it slides is  63N.

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

The friction force is given by

f = coefficient of friction x Normal force

Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction N =35 × 9.8 = 343N

Frictional force f =0.4 x 343

                          f =137.20N

The net force will be

Fnet= Fappplied - Ffrictional

Fnet = 200-137.20 = 62.8N

Fnet = 63N

Thus,  the net force on the couch as it slides is  63N.

Learn more about friction force.

https://brainly.com/question/1714663

#SPJ2

Answer:

a population increase

Explanation:

During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.

Answer:

(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center

(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.

Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.

(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.

Answer:

(a) f = 185 Hz

(b) v = 266.4 m/s

Explanation:

(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:

[tex]f_n=\frac{nv}{2L}[/tex]

[tex]f_n=nf[/tex]

n: order of the mode

v: velocity of the waves in the string

L: length of the string = 72.0cm = 0.72m

fn: frequency of the n-th mode

With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:

[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]

you solve the previous equation for n:

[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]

With this information you can calculate the lowest resonant frequency:

[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]

b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:

[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]

fn = 555 Hz

fn-1: 370 Hz

[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´

hence, the velocityof the waves in the string is 266.4 m/s

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

Answer:

In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.

Explanation:

Hope this helps!

Answer:

1.48 g

Explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

A = (5.00 g) (½)^(5.00 a / 2.85 a)

A = 1.48 g

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

Answer: 1.00c

Explanation: I got it correct on the homework

Answer:

The distance covered by puck A before collision is  [tex]z = 8.56 \ m[/tex]

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  [tex]v_A = 3.90 \ m/s[/tex]

        The speed of puck B is  [tex]v_B = 4.30 \ m/s[/tex]

The distance covered by puck A is mathematically represented as

     [tex]z = v_A * t[/tex]

  =>  [tex]t = \frac{z}{v_A}[/tex]

 The distance covered by puck B  is  mathematically represented as

      [tex]18 - z = v_B * t[/tex]

=>   [tex]t = \frac{18 - z}{v_B}[/tex]

Since the time take before collision is the same

        [tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]

substituting values

          [tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]

=>      [tex]70.2 - 3.90 z = 4.3 z[/tex]

=>       [tex]z = 8.56 \ m[/tex]

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, [tex]T_s[/tex] = 5,300 K

temperature of the Earth, [tex]T_E[/tex] = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]

The earth gains Q of heat and therefore increases its entropy by the amount

[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]

The total entropy change is:

[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

Answer:

Exothermic

Explanation:

Depending on the unit you are in, the answer may vary.

This is an exothermic reaction because it produces heat (the beaker gets warm).

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

a = 1.72 m/s²

Answer:

The speed of the freight car decreases.

Explanation:

According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received

In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.

Therefore the speed of the freight car decreased by applying the law of conservation of momentum i

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

r = 20 m