In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.

Answers

Answer 1

Answer:

1.24Mev

Explanation:

Using

E= hc/lambda

= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9

= 1.24Mev


Related Questions

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed:_____

a. into the page.
b. toward the left
c. toward the right
d. toward the bottom of the page.
e. toward the top of the page.
f. out of the page.

Answers

Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,

1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N

Answers

Explanation:

Pressure is the same for both plungers.

P = P

F / A = F / A

F / (¼ π d²) = F / (¼ π d²)

F / d² = F / d²

5 N / (0.05 m)² = F / (1 m)²

F = 2000 N

None of the options are correct.

PLEASE HELP WILL GIVE BRAINLIEST In an experiment, the hypothesis is that if leaf color is related to temperature, then exposing the plant to low temperatures will result in a leaf color change . This hypothesis is _____. 1. testable 2.falsifiable 3.a and b above 4.none of the above

Answers

Answer:

testable

Explanation:

high heat can cause browning and or welting

Answer:

A

Explanation:

If it is tested, it will really changed when the plant into low temperature

light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

Answers

Answer:

121ohms

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

Hence, the resistance of this bulb is 121 ohms

Which of the following statements is true vibrations ?

Answers

Answer:

C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.

Explanation:

The complete question is

Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.

Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.

Answer:

Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."

Explanation:

My options were:

A)  Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.

B)  Resonance occurs as a result of sympathetic vibrations.

C)  A non-vibrating object can begin to vibrate as a result of forced vibrations.

D)  Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.

A is correct

.

A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g

Answers

Answer:

1,860 g

Explanation:

In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular reflection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort of diffuse reflection.

Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an angle alpha (from the horizontal) and one inclined at an angle beta. What would the angular separation between the rays reflected from the two surfaces?

Answers

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 275 V. Assume a plate separation of d 1.53 cm and a plate area of A = 25.0 cm2. when the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
(a) Calculate the charge on the plates in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
before Qi = _____
after Qf = ______
(b) Determine the capacitance (in F) and potential difference (in V) after immersion
(c) Determine the change in energy (in n]) of the capacitor Δυ = nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 275 V potential difference
Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.)
Determine the capacitance (in F) and potential difference (in V) after immersion
Determine the change in energy (in nJ) of the capacitor AU nJ

Answers

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.

(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.

(c) The change in energy of the capacitor is 54.02 nJ.

Charge on the plate before immersion

The charge on the plate is calculated as follows;

[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]

Charge on the plate after immersion

[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]

Capacitance and potential difference after immersion

[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]

[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]

Change in energy of the capacitor

The initial energy of the capacitor is calculated as follows;

[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]

The final energy of the capacitor is calculated as follows;

[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]

Change in energy is calculated as follows;

[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]

Learn more about energy stored in a capacitor here: https://brainly.com/question/13578522

g A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75 C/s . Part A How many electrons are delivered to the

Answers

Answer:

Approximately 5 x 10^18 electrons are delivered to the smart phone charger.

Explanation:

The electric current in a circuit is the flow of charges through a circuit with time.

The charges through the circuit are due to the electrons that flow through the circuit.

An individual electrons has a charge of -1.60 x 10^-19 C on it.

If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.

==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.

If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g

Answers

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current [tex]I_1[/tex] = 100 μA

current [tex]I_2[/tex] = 1 mA

forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode  if the bias current is increased to 1mA.

Using the formula:

[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

where;

[tex]V_r[/tex] = 0.7

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]

[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

Suppose n = 1

[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]

Then;

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]

[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]

[tex]V_r'=29.37 \times nV_T[/tex]

[tex]V_r'=29.37 \times 25.86[/tex]

[tex]V_r'=759.5 \ mV[/tex]

[tex]Vr' \simeq[/tex] 760 mV

Thus, the voltage drop across this same diode will be 760 mV

The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge?

Answers

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

q = 180000(9×10⁹)/0.028²

q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?

Answers

Answer:

a

    [tex]z= 2.5 \ m[/tex]

b

   [tex]z = (1 \ m , 4 \ m )[/tex]

Explanation:

From the question we are told that

     Their distance apart is  [tex]d = 5.00 \ m[/tex]

      The  wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              [tex]z - (d-z) = m \lambda[/tex]

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        [tex]z - (5-z) = 0[/tex]

=>     [tex]2 z - 5 = 0[/tex]

=>     [tex]z= 2.5 \ m[/tex]

Generally the path difference for destructive  interference is

           [tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]

=>         [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]

=>        [tex]|2z - d| =\frac{\lambda}{2}[/tex]

substituting values

          [tex]|2z - 5| =\frac{6}{2}[/tex]

=>      [tex]z = \frac{5 \pm 3}{2}[/tex]

So  

      [tex]z = \frac{5 + 3}{2}[/tex]

      [tex]z = 4\ m[/tex]

and

      [tex]z = \frac{ 5 -3 }{2}[/tex]

=>   [tex]z = 1 \ m[/tex]

=>    [tex]z = (1 \ m , 4 \ m )[/tex]

The magnitude of the magnetic field at point P for a certain electromagnetic wave is 2.12 μT. What is the magnitude of the electric field for that wave at P? (c = 3.0 × 108 m/s)

Answers

Answer:

The electric field is  [tex]E = 636 \ V/m[/tex]

Explanation:

From the question we are told that

     The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]

      The value for speed of light is  [tex]c = 3.0 *10^8 \ m/s[/tex]

Generally the magnitude of the electric field at point P is

        [tex]E = B * c[/tex]

substituting values

         [tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]

         [tex]E = 636 \ V/m[/tex]

The magnitude of electric field for the wave at point P is 636 V/m.

Given data:

The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].

The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].

The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,

[tex]E = B \times c[/tex]

here,

E is the strength of electric field.

Solving as,

[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]

Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.

Learn more about the electric field here:

https://brainly.com/question/15800304

Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.75
Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.23×1030kg
. Find the radius of the exoplanet's orbit.

Answers

Answer:

[tex]r=4.24\times 10^{11}\ m[/tex]

Explanation:

Given that,

Orbital time period, T = 3.75 earth years

Mass of star, [tex]m=3.23\times 10^{30}\ kg[/tex]

We need to find the radius of the exoplanet's orbit. It is a concept of Kepler's third law of motion i.e.

[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]

r is the radius of the exoplanet's orbit.

Solving for r we get :

[tex]r=(\dfrac{T^2GM}{4\pi^2})^{1/3}[/tex]

We know that, [tex]1\ \text{earth year}=3.154\times 10^7\ \text{s}[/tex]

So,

[tex]r=(\dfrac{(3.75\times 3.154\times 10^7)^2\times 6.67\times 10^{-11}\times 3.23\times 10^{30}}{4\pi^2})^{1/3}\\\\r=4.24\times 10^{11}\ m[/tex]

So, the radius of the exoplanet's orbit is [tex]4.24\times 10^{11}\ m[/tex].

Monochromatic light falls on two very narrow slits 0.047 mm apart. Successive fringes on a screen 6.60 m away are 8.9 cm apart near the center of the pattern.
Determine the wavelength and frequency of the light.

Answers

Answer::

   [tex]\lambda = 634 nm[/tex]

  [tex]f = 4.73 *10^{14} \ Hz[/tex]

Explanation:

From the question we are told that

      The  distance of separation is  [tex]d = 0.047 \ mm = 0.047 *10^{-3} \ m[/tex]

       The  distance of the screen is  [tex]D = 6.60 \ m[/tex]

      The  width of the fringe is [tex]y = 8.9 \ cm = 0.089 \ m[/tex]

     

Generally the width of the width of the fringes is mathematically represented as

          [tex]y = \frac{\lambda * D }{d }[/tex]

=>       [tex]\lambda = \frac{y * d }{D }[/tex]

=>      [tex]\lambda = \frac{ 0.089 * (0.047 *10^{-3}) }{6.60 }[/tex]

=>    [tex]\lambda = 634 *10^{-9}[/tex]

=>  [tex]\lambda = 634 nm[/tex]

Generally the speed of light is mathematically represented as

         [tex]c = f * \lambda[/tex]

=>       [tex]f= \frac{ c}{\lambda }[/tex]

=>         [tex]f= \frac{ 3.0 *10^{8}}{634 *10^{-9}}[/tex]

=>     [tex]f = 4.73 *10^{14} \ Hz[/tex]

A mass m = 0.7 kg is released from rest at the origin 0. The mass falls under the influence of gravity. When the mass reaches point A, it is a distance x below the origin 0; when the mass reaches point B it is a distance of 3 x below the origin 0. What is vB/vA?

Answers

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Which of the following frequencies could NOT be present as a standing wave in a 2m long organ pipe open at both ends? The fundamental frequency is 85 Hz.

Answers

Answer:

382Hz

Explanation:

The question lacks the required option. Find the complete question in the attachment.

The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L

Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...

Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;

First overtone f1 = 2fo = 2(85) = 170Hz

Second overtone f2 = 3fo = 3(85) = 255Hz

Third overtone = 4fo = 4(85) = 340Hz

Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz

"In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to"

Answers

Answer:

To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D

A steel ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring constant is 6.00 N/m. When the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 26.1 cm/s. (a) What is the mass of the ball (in kg)? kg (b) What is the period of oscillation (in s)? s (c) What is the maximum acceleration of the ball? (Enter the magnitude in m/s2.) m/s2

Answers

Answer:

a) m = 0.626 kg , b) T = 2.09 s , c)   a = 1.0544 m / s²

Explanation:

In a spring mass system the equation of motion is

        x = A cos (wt + Ф)

with      w = √(k / m)

a) velocity is defined by

        v = dx / dt

        v = - A w sin (wt + Ф)          (1)

give us that the speed is

        v = 26.1 m / s

for the point

        x = a / 2

the range of motion is a = 11.0 cm

       x = 11.0 / 2

       x = 5.5 cm

Let's find the time it takes to get to this distance

       wt + Ф = cos⁻¹ (x / A)

       wt + Ф = cos 0.5

        wt + Ф = 0.877

In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f

       Ф = 0

       wt = 0.877

       t = 0.877 / w

we substitute in equation 1

       26.1 = -11.0 w sin (w 0.877 / w)

        w = 26.1 / (11 sin 0.877))

        w = 3.096 rad / s

from the angular velocity equation

       w² = k / m

       m = k / w²

       m = 6 / 3,096²

       m = 0.626 kg

b) angular velocity and frequency are related

       w = 2π f

frequency and period are related

        f = 1 / T

we substitute

        w = 2π / T

        T = 2π / w

        T = 2π / 3,096

        T = 2.09 s

c) maximum acceleration

 the acceleration of defined by

        a = dv / dt

        a = - Aw² cos (wt)

the acceleration is maximum when the cosine is ±1

         a = A w²

          a = 11  3,096²

        a = 105.44 cm / s²

we reduce to m / s

        a = 1.0544 m / s²

Find the angle in degrees for the third-order maximum for 577 nm wavelength yellow light falling on a diffraction grating having 1,420 lines per centimeter.

Answers

Answer:

θ = 0.14°

Explanation:

Here we will use the grating equation. The grating equation is as follows:

mλ = d Sin θ

where,

θ = angle = ?

m = order number = 3

λ = wavelength of light = 577 nm = 5.77 x 10⁻⁷ m

d = spacing between slits = 1/(1420 lines/cm) = 7.042 x 10⁻⁴ m

Therefore, using these values, we get:

(3)(5.77 x 10⁻⁷ m) = (7.042 x 10⁻⁴ m)Sin θ

Sin θ = (3)(5.77 x 10⁻⁷ m)/(7.042 x 10⁻⁴ m)

Sin θ = 2.46 x 10⁻³

θ = Sin⁻¹(2.46 x 10⁻³)

θ = 0.14°

A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?

Answers

Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Mention the importance of writing the physical quantities as vectors.​

Answers

Answer:

Knowledge of vectors is important because many quantities used in physics are vectors. If you try to add together vector quantities without taking into account their direction you'll get results that are incorrect.

Explanation:

An example of the importance of vector addition could be the following:

Two cars are involved in a collision. At the time of the collision car A was travelling at 40 mph, car B was travelling at 60 mph. Until I tell you in which directions the cars were travelling you don't know how serious the collision was.

The cars could have been travelling in the same direction, in which case car B crashed into the back of car A, and the relative velocity between them was 20 mph. Or the cars could have been travelling in opposite directions, in which case it was a head on collision with a relative velocity between the cars of 100 mph!

for an answer to be complete,the units needs to be specified.why

Answers

Explanation:

unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given

A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?

Answers

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]

Thus, the effective dose allowable yearly = 16 times

Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the collision. What is the wavelength of the scattered photon?

Answers

0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 2 cm

Answers

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is the angle of refraction in degrees?

Answers

Answer:

   θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

        n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

         sin θ₂ = n₁ / n₂ sin θ

let's calculate

         sin = 1 / 1.3 sin 0.23

         sin = 0.175

        θ= 0.17528 rad

let's reduce to degrees

       θ = 0.17528 rad (180ª / pi rad)

       θ = 10.28º

If a train travels at a constant 18.0 m/s, how far would it move in one hour? In 1.00 minute? In 1.00 second?

Answers

Explanation:

Distance = speed × time

d = (18.0 m/s) (1 hr × 3600 s/hr)

d = 64,800 m

d = (18.0 m/s) (1 min × 60 s/min)

d = 1080 m

d = (18.0 m/s) (1 s)

d = 18.0 m

Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body position to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.

Answers

Answer:

Option C - she will speed up because air resistance has reduced and be less than gravity

Explanation:

We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body

Since the downward gravitational force on Luz is constant, she will fall with a net force of;

F_net = F_g - F_d

where;

F_net is the net force on Luz acting downwards

F_g is the gravitational force on Luz

F_d is the drag force on Luz

The drag force on her body is proportional to the surface area of attack.

We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.

Now, from earlier, we saw that;

F_net = F_g - F_d

So, the lesser F_d is, the higher F_net becomes.

Thus, she will speed up because air resistance has reduced and be less than gravity.

Answer:

C

Explanation:

EDGE 2020

¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?

Answers

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

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