Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced

Answers

Answer 1

Answer:

4.635 *10^12 Neutrinos

Explanation:

Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.

We proceed as follows;

In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.

From the question, we are given that

Power P = 9.9 watts = 9.9 J/s

Watts is same as J/s

The number of proton-proton cycles required to generate E energy is N = E / E '

Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question

Converting Mev to J, we have

= 26.7 x1.6 x10 -13 J

To get the number N which is the number of proton-proton cycle required, we have;

N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12

Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.

Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos


Related Questions

In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to easily count them. To spread out the fringe pattern, one could

Answers

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

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

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration

Answers

Answer:

The acceleration of the refrigerator together with the objects decreases.

Explanation:

If the mass of the refrigerator is increased by stacking more masses (objects) on it,

and the force applied remains constant, then we know from

F = ma

where

F is the applied force

m is the total mass of the refrigerator and the objects

a is the acceleration of the masses.

If F is constant, and m is increased, the acceleration will decrease

Answer:

The acceleration decreases.

Explanation:

its right

(4) Use the preliminary observations to answer these questions; Compared to no polarizer or analyzer in the optical path, by what percent does the light intensity decrease when (a) The polarizer is introduced into the optical path? (b) The both polarizer and analyzer are introduced into the optical path?

Answers

Answer:

a)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.

Answers

Answer:

a) k = 95.54 N / m,   c =   19.55 , b)      m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

          fr = -c v

when solving the system the angular value remains

          w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

     f₁ = 1.1 Hz

the angular velocity is related to frequency

         w = 2π f₁

Let's find the angular velocity without friction is

         w₂ = k / m₁

we substitute

        (2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

    f₂ = 0.8 Hz

        (2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

           (2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

           k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

           k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

           k = 4 π² (0.8² -1.1²) / (½ -1/1)

           k = 39.4784 (1.21) / (-0.5)

           k = 95.54 N / m

now we can find the constant of friction

              (2π f₁) 2 = k / m₁ + (c / 2m₁)²

           c2 = ((2π f₁)² - k / m₁) 4m₁²

           c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

           c² = (4π² 1,1² - 95,54 / 1) 4 1²

           c² = (47.768885 - 95.54) 8

           c² = -382.1689

           c =   19.55    

b) f₃ = 0.2 Hz

   m₃ =?

              (2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

              (4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

              1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

              x = 1 / m₃

              x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

              60.5080 x² + 60.5080 x -1 = 0

                x² + x - 1.65 10⁻² = 0

                  x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

                  x = [1 ± 1.03] / 2

                  x₁ = 1.015 kg

                  x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

                  x₁ = 1 / m₃

                 m₃ = 1 / x₁

                  m₃ = 1 / 1.1015

             

An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Answers

Answer:

Current, I = 1.23 A

Explanation:

Given that,

Inductance, L = 35 mH

Resistance, R = 12 ohms

Potential difference, V = 18 V

We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :

[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)

Here,

[tex]I_o[/tex] is final current

[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]

[tex]\tau[/tex] is time constant

[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]

So, equation (1) becomes :

[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]

So, after 5 ms the current in the circuit is 1.23 A.

Why was Bohr's atomic model replaced by the
modern atomic model?

Answers

Answer:

Explanation:

Bohr's atomic model was replaced by the  modern atomic model because of its limitations, which included :

(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )

(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).

(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.

A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth

Answers

Answer:

t₀ = 1.55 s

Explanation:

According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in  motion relative to the event is not the same as measured by an observer at rest.

It is given as:

t = t₀/[√(1 - v²/c²)]

where,

t = time measured by astronaut in motion = 3.1 s

t₀ = time required according to observer on earth = ?

v = relative velocity = 0.87 c

c = speed of light

3.1 s = t₀/[√(1 - 0.87²c²/c²)]

(3.1 s)(0.5) = t₀

t₀ = 1.55 s

Answer:

The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]

Explanation:

Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]

where,

[tex]t_o = 3.1\\\\v = 0.87[/tex]

[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]

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A planar electromagnetic wave is propagating in the x direction. At a certain point P and at a given instant, the magnitude of the electric field of the wave is 0.082 V/m . What is the magnetic vector of the wave at point P at that instant?
A) (0.27 nT)k
B) (-0.27 nT)k
C) (0.27 nTİ
D) (6.8 nT)k
E) (-6.8 nT))

Answers

Answer:

b

Explanation:

A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 10-7 N/A2 .) A. 0.02219 m B. 327 m C. 52 m D. 0.00199 m

Answers

Answer:

A. 2.2*10^-2m

Explanation:

Using

Area = length x L/ uo xN²

So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*

3000²)

A = 17.5*10^-3/ 1.13*10^-5

= 15.5*10^-2m²

Area= π r ²

15.5E-2/3.142 = r²

2.2*10^2m

Explanation:

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion

Answers

Answer:

Explanation:i think this would help u

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm

Answers

Answer:

Explanation:

In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is

2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .

2 x 1.34 x t = 1  x 580

t = 216.42 nm .

Thickness must be 216.42 nm .

You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?

Answers

Answer:

a) Connect a series resistance of 8,47 ohms

b)14,16 [A]

c) r = 10,96 ohms

Explanation:

My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower

In America

P = V*I

1700 (w) = 120*I

I = 1700/120 [A]

I = 14,16 [A]        current needed for the blower

In Europe

120 (v)  (the drop of voltage I need) when a current of 14,16 passes through to series  resistor is

V = I*R          120 = 14,16* R         R = 8,47 ohms

c) P = I*r²

1700 (w) = 14,16 (A) * r²

r² = 120,06

r = 10,96 ohms

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?

Answers

Answer:

200cm

Explanation:

Answer:

100cm

Explanation:

Using

F= ( N/2L)(√T/u)

F1 will now be (0.5*2)( √600/0.015)

=> L( wavelength)= 200/2cm = 100cm

A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?

Answers

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to

Answers

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen

Answers

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m


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In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.

Answers

Answer:

Explanation:

We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.

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A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy

Answers

Answer: her rotational kinetic energy increases

A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.

Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.

Answers

Answer:

The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)  

Explanation:

Wavelength

Convert all lengths to meters:

Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].

Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:

The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.

These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].

The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)

Therefore:

[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].

On the other hand:

[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].

Because the cotangent of [tex]\theta[/tex] is very close to zero,

[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].

[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].

[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].

Distance between two adjacent maxima

If the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.

The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].

On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:

[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].

Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].

A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?

Answers

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]

So it would normally be written with a negative signs in front indicating a divergent lens.

What happens to the magnetic field when you reverse the direction of current by sliding the battery voltage bar past 0 volts

Answers

Answer:

The polarity of the magnetic field changes

Explanation:

This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.

Answers

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor

Answers

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]

[tex]0.8 = e^{-\frac{88}{R} }[/tex]

[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]

[tex]ln(0.8) = -\frac{88}{R}[/tex]

[tex]R = -\frac{88}{ln(0.8)}[/tex]

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.

Answers

The correct answer is C. The inner planets are also called terrestrial planets.

Explanation:

Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".

Answer:

The answer is c.) The inner planets are also called terrestrial planets.

Explanation:

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?

Answers

Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.

              Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m

Explanation: A speed of wave on a string under a tension force can be calculated as:

[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]

[tex]F_{T}[/tex] is tension force (N)

μ is linear density (kg/m)

Determining velocity:

[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{0.00874 }[/tex]

[tex]|v| =[/tex] 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]

Δx = 11.5×[tex]10^{-6}[/tex]

With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex]  m.

Doubling Tension:

[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]

[tex]|v| = \sqrt{2.0.00874 }[/tex]

[tex]|v| = \sqrt{0.01568}[/tex]

|v| = 0.1252 m/s

Displacement for same time:

[tex]\Delta x = |v|.t[/tex]

[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]

[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]

With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answers

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.610 T.Required:a. What orientation of the coil gives the maximum torque on the coil ?b. What is this maximum torque in part (A) ?c. For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)?

Answers

Answer:

a) for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

b) therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

c) Given the torque is 71.0% of its maximum value; Ф  = 45.24⁰ ≈ 45⁰

Explanation:

Given that; Diameter is 8.40 cm,

Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m

Number of turns (N) = 17

Current in the loop (I) = 3.20 A

Magnetic field (B) = 0.610 T

Let the angle between the loop's area vector A and the magnetic field B be

Now. the area of the loop is;

A = πR²

A = 3.14 ( 0.042 )²

A =  0.005539 m²

Torque on the loop (t) = NIABsinФ

t = 17 × 3.20 ×0.005539 × 0.610 × sinФ

t = 0.1838sinФ N.m

for the torque to be maximum, sin should be maximum

i.e (sinФ)maximum = 1

therefore the Maximum torque is

Tmax = 0.1838 × 1 = 0.1838  N.m

Given the torque is 71.0% of its maximum value

t = 0.71 × tmax

t = 0.71 × 0.1838

t = 0.1305

Now

0.1305 N.m =  0.1838 sinФ N.m

sinФ = 0.1305 / 0.1838

sinФ = 0.71001

Ф = sin⁻¹ 0.71001

Ф  = 45.24⁰ ≈ 45⁰

Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).

Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

Answers

Answer:

a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².

Explanation:

a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.

τ = Frsinθ

= 185 N × 1.73 m × sin90°

= 320.05 Nm

So, the torque τ = 320.05 Nm

b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m

α = angular acceleration of grindstone

τ = Iα

α = τ/I = τ/(MR²/2) = 2τ/MR²

substituting the values of the variables, we have

α = 2τ/MR²

= 2 × 320.05 Nm/[76 kg × (1.73 m)²]

= 640.1 Nm/227.4604 kgm²

= 2.814 rad/s²

So, the angular acceleration α = 2.814 rad/s²

c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.

So  τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm

The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm

Since τ'' = Iα

α' = τ''/I where α' = its new angular acceleration

α' = 2τ/MR²

= 2 × 319.75 Nm/[76 kg × (1.73 m)²]

= 639.5 Nm/227.4604 kgm²

= 2.811 rad/s²

So, the angular acceleration α' = 2.811 rad/s²

n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How fast does the electron move away from the metal

Answers

Answer:

The speed of the electron is 1.371 x 10 m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]

Therefore, the speed of the electron is 1.371 x 10 m/s.

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