A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in the opposite directions at 0.45 m/s. Find the child's mass.

The bug moving toward the center of the disk changes the moment of inertia and the angular velocity, resulting in a change in the rotational kinetic energy of the system. The principles of conservation of angular momentum and calculations of kinetic energy apply here.

This problem involves the principle of conservation of angular momentum. Initially, when the bug is at the edge of the disk, the system's angular momentum (L) is conserved. The initial angular momentum (L_initial) is the sum of the angular momenta of the bug and the disk. It can be calculated as L_initial = I_bug(ω) + I_disk(ω) where I_bug = m*R², I_disk = (1/2)*M*R², m is the mass of the bug, M is the mass of the disk, R is the radius of the disk, and ω is the initial angular velocity.

When the bug moves to the center of the disk, the angular velocity will change but the total angular momentum will still remain conserved as L_final = I_bug(ω') + I_disk(ω') where ω' is the final angular velocity and I_bug is now 0 because the bug is at the center. From the conservation of angular momentum, we can solve for ω'.

The change in kinetic energy can be calculated from the difference between the initial kinetic energy and the final kinetic energy of the system. An increase or decrease in kinetic energy in this scenario is due to the displacement of the bug from the edge to the center of the disk, which changes the moment of inertia of the system and thus, the rotational kinetic energy.

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Final answer:

The explanation covers the new angular velocity of the disk, the change in kinetic energy, and the causes for the increase and decrease in kinetic energy.

Explanation:

A: The new angular velocity of the disk can be calculated using the principle of conservation of angular momentum. The bug crawling towards the center decreases the moment of inertia, resulting in an increase in angular velocity.
B: The change in kinetic energy of the system can be determined by calculating the initial and final kinetic energies and finding their difference.
C: The increase in kinetic energy is due to the bug moving towards the center, reducing the moment of inertia. The decrease in kinetic energy occurs due to the redistribution of mass towards the center of rotation.

Explanation:

We have,

Slit separation, d = 0.329 mm

Distance between slit and screen, D = 2.20 m

The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.

For double slit experiment, the fringe width is given by :

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

[tex]\lambda[/tex] is wavelength

[tex]\lambda=\dfrac{\beta d}{D}\\\\\lambda=\dfrac{2.9\times 10^{-3}\times 0.329\times 10^{-3}}{2.2}\\\\\lambda=4.33\times 10^{-7}\ m\\\\\lambda=433\ nm[/tex]

So, wavelength is 433 nm.

Answer:

E = 3,154.37 KJ

The heat of combustion per mole of pentane is 3,154.37 KJ

Explanation:

Given;

Change in temperature of system ∆T = 23.65-20.45 = 3.2 °C

Mass of water m1 = 1 kg

Specific heat capacity of water C1 = 4.184J/g°C = 4184J/kg °C

Heat capacity of calorimeter mC2 = 2.21 kJ/°C = 2210J/°C

Heat gained by both calorimeter and water is;

H = (m1C1 + mC2)∆T

Substituting the values;

H = (1×4184 + 2210)×3.2

H = 20460.8 J

Mass of pentane burned = 0.468 g

Molecular mass of pentane = 72.15g

If 0.468g of pentane releases 20460.8 J of heat,

72.15g of pentane will release;

E = (72.15/0.468) × 20460.8 J

E = 3154373.333333J

E = 3,154.37 KJ

The heat of combustion per mole of pentane is 3,154.37 KJ

Answer:

Explanation:

capacitance = 20 x 10⁻⁶ F .

potential V = 12 V

charge = CV

= 20 x 10⁻⁶ x 12

Q = 240 x 10⁻⁶ C

energy = Q² / 2C

= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶

= 1440 x 10⁻⁶ J

b )

In this case charge will remain the same but capacity will be increased 4 times

new capacity C = 4 x 20 x 10⁻⁶

= 80 x 10⁻⁶

energy = Q² / 2C

=  (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶

= 360 x 10⁻⁶ J .

potential energy will decrease from  1440 x 10⁻⁶ J to 360 x 10⁻⁶ J

Answer:

Explanation:

length of wire is proportional to frquency of sound produced by it.

n₁ /n₂ = l₂ / l₁

= 21 / 20

If n be the frequency of tuning fork

n₁ = n + 5  ;  n₂ = n - 5   ( no of beat / s  = frequency difference )

n + 5 / n - 5 = 21 / 20

20n + 100 = 21n - 105

n = 205 Hz

frequency of tuning fork = 205

the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm.   = n - 5 = 200 Hz .

Given the beat frequency and the inverse relationship between the length of the sonometer wire and frequency, we determine the tuning fork frequency. Beat frequency remains constant despite the change in wire length. Calculation involves solving system of linear equations with fundamentals of wave physics.

To find the frequency of the tuning fork and the frequency of the sonometer wire for both lengths, we utilize the concept of beat frequency. The beat frequency is given by the difference in frequency between the tuning fork and the sonometer wire.

Let’s denote the frequency of the tuning fork as ft and the frequency of the sonometer wire for 20 cm as f1. Given that the beat frequency is 5 Hz, we have:

|ft - f1| = 5 Hz

When the length of the wire is changed to 21 cm, assume the new frequency of the sonometer wire to be f2. Beat frequency remains unchanged:

|ft - f2| = 5 Hz

The frequency of a vibrating string is inversely proportional to its length:

f1 * L1 = f2 * L2

Given L1 = 20 cm and L2 = 21 cm,

f1 * 20 = f2 * 21 which simplifies to:

f2 = (20/21) * f1

Since |ft - f1| = 5 and |ft - f2| = 5, solving these equations requires knowing the possible frequencies of the sonometer.

Determine f1 and f2 using the equations:

Solve both scenarios for ft + 5 = f1 and ft - 5 = f2:

If ft= f1 ± 5 Hz,

Scenario 1: for f1 = (n/20), f2 = (20/21) * (n/20) = n/21, verify both.

Elements in the same group on the periodic table, like alkali metals lithium and sodium or alkaline earth metals beryllium and magnesium, have the same number of valence electrons. This similarity is a key factor in their chemical reactivity and the formation of covalent bonds, evidenced through electron dot diagrams.

Elements that have the same number of valence electrons are located in the same group or column on the periodic table. For instance, the alkali metals such as lithium (Li) and sodium (Na) each possess one valence electron. Similarly, the alkaline earth metals like beryllium (Be) and magnesium (Mg) each have two valence electrons.

Another example can be found in the halogens group, as elements such as fluorine (F) and chlorine (Cl) each exhibit seven valence electrons. These elements have not only the same valence electron count but also exhibit similar chemical properties due to this shared characteristic. It's important to note that the number of valence electrons contributes to an element's ability to bond with others through the loss, gain, or sharing of these electrons.

Understanding that elements in the same group on the periodic table share the same number of valence electrons can greatly aid in predicting their chemical behavior and reactivity. Electron dot diagrams provide a visual representation of the valence electron distribution in each element and are particularly useful for visualizing how elements bond covalently. For example, elements in the first group have a single dot representing the single valence electron they possess.

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude of force is  [tex]F_R} = 5.33 \ m N[/tex]

Explanation:

 From the question we are told that

          The current in wire A , B and C are   [tex]I _a = I_b =I_c = I= 20 A[/tex]

          The distance is [tex]R = 15.0mm = \frac{15}{1000} = 0.015m[/tex]

           The length of wire C is [tex]L_c = 200cm = \frac{200}{100} = 2 m[/tex]

Generally the force exerted per unit length  that is acting in between two current carrying conductors can be mathematically represented as

             [tex]\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}[/tex]

   Where [tex]\mu_o[/tex] is the permeability  of free space with a constant value of  

              [tex]\mu_o = 4 \pi * 10^{-7} N/A^2[/tex]

When the current in the wire are of the same direction  the force is positive and when they are in opposite direction the force is negative

   Considering force between A and C

          [tex]F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L[/tex]

   Considering force between B and C    

        [tex]F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]

The resultant force is

        [tex]F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]

         [tex]F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })[/tex]

Substituting values

       [tex]F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })[/tex]

       [tex]F_R} = 5.33 *10^{-3} N[/tex]

       [tex]F_R} = 5.33 \ m N[/tex]

Answer:

See explaination

Explanation:

Magnetic flux density definition, a vector quantity used as a measure of a magnetic field.

It can be defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction.

See attached file for detailed solution of the given problem.

The magnetic flux density inside a solenoid can be derived using the formula B = μ₀(NI)/L. As L approaches infinity, this simplifies to B = μ₀nI, aligning with the standard magnetic field expression for a long solenoid.

To find the magnetic flux density at a point on the axis of a solenoid with radius b, length L, a current I, and N turns of closely wound coil, use the following steps:

Step 1 : Consider an infinitesimal element dz of the solenoid. The element is located at a distance z from the center of the solenoid. The magnetic field contribution ( dB ) from this element at point ( P ) is given by the Biot-Savart law as -

[tex]\[ dB = \frac{\mu_0 I d\ell}{4\pi} \frac{1}{r^2} \][/tex]

Step 2 : For a solenoid, considering a differential element, the distance r from the element to the point P on the axis is -

[tex]\( \sqrt{b^2 + (z - x)^2} \).[/tex]

Step 3 : For a small segment dz, the current element is -

[tex]\[ dB = \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Step 4 :  Integrate dB from -L/2 to L/2 :

[tex]\[ B = \int_{-L/2}^{L/2} \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Substitute, [tex]\( n = \frac{N}{L}[/tex]

[tex]\[ B = \frac{\mu_0 I N}{4\pi L} \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Solving the integral part -

[tex]\[ \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} = \frac{z}{b^2 \sqrt{b^2 + (z - x)^2}} \Bigg|_{-L/2}^{L/2} \][/tex]

[tex]\[ \left[ \frac{L/2}{b^2 \sqrt{b^2 + (L/2 - x)^2}} - \frac{-L/2}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \right] \][/tex]

Step 5: When  L  is very large, the ends of the solenoid are far away, and we can approximate:

[tex]\[ \frac{L/2 - x}{b^2 \sqrt{b^2 + (L/2 - x)^2}} \approx \frac{1}{b^2} \quad \text{and} \quad \frac{-L/2 - x}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \approx -\frac{1}{b^2} \][/tex]

This simplifies the result to -

[tex]\[ \left[ \frac{L/2 - x}{b^2 (L/2 - x)} - \frac{-L/2 - x}{b^2 (L/2 + x)} \right] = \left[ \frac{1}{b^2} - \left(-\frac{1}{b^2}\right) \right] = \frac{2}{b^2} \][/tex]

Conclusion, For a very long solenoid:

[tex]\[B = \mu_0 \frac{N}{L} I\][/tex]

Which simplifies to,

[tex]\[B = \mu_0 n I\][/tex]

Thus, for a very long soleniod the magnetic flux density can be approximated to [tex]\[B = \mu_0 n I\][/tex].

Answer: Black hole.

Explanation:

As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.

Answer:

E = 301.5 J

Explanation:

We have,

Mass of mother, m = 67 kg

Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.

It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.

The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2[/tex]

So, the elastic potential energy is :

[tex]E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J[/tex]

So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.

An observer on Starship B will measure the same time interval as that on Starship A, because they are both travelling at the same speed and in the same direction relative to Earth. Starship C's measurement would differ due to its opposite direction of travel relative to A and B.

To determine which reference frame measures a time interval equal to the time interval measured by the clock aboard Starship A, we need to consider the principles of the theory of special relativity. Since Starship B is traveling at the same speed and in the same direction as Starship A relative to Earth, the time dilation effect will be the same for both starships. Therefore, an observer on Starship B will measure the same time interval on their clock as that measured on Starship A's clock.

In contrast, Starship C is moving at the same speed but in the opposite direction relative to Earth, and its relative velocity to Starships A and B is not zero. As a result, the time interval measured on Starship C will not match the time interval measured on Starship A's clock.

So, the time interval that equals the time interval measured by the clock aboard Starship A can be measured in the reference frame of Starship B, which is moving at the same speed and in the same direction as Starship A relative to Earth.

Answer:

a) 86 atm

b) 86 atm

c) 645 m/s

Explanation:

See attachment for calculations on how i arrived at the answer

The atmospheric pressure of 2.00 km above the surface of Venus is 86 atm and the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km is 645 m/sec.

Given :

a) and b) In order to determine the atmospheric pressure of 2.00 km above the surface of Venus using the formula given below:

[tex]\rm P = P_0\times L^{\frac{Mg}{\rho T}}[/tex]   --- (1)

The value of the expression [tex]\rm Mg/\rho T[/tex] is given below:

[tex]\rm \dfrac{Mg}{\rho T} = \dfrac{44\times 10^{-3}\times 9.8\times 10^3}{8.314\times 733}[/tex]

[tex]\rm \dfrac{Mg}{\rho T}=0.02076[/tex]

Now, substitute the values of the known terms in the expression (1).

[tex]\rm P=92\times L^{0.02076}[/tex]

P = 86 atm

c) The root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km can be calculated as given below:

[tex]\rm V_{rms}=\sqrt{\dfrac{3RT}{M}}[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\rm V_{rms}=\sqrt{\dfrac{3\times 8.314\times 733}{44\times 10^{-3}}}[/tex]

Simplify the above expression.

[tex]\rm V_{rms} = 645\;m/sec[/tex]

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Answer:  Increase the amount of carbon dioxide in the ecosystem.

Explanation:

The producers are the organisms which can make their own food by conducting the process of photosynthesis. The carbon dioxide, water are the reactants and carbohydrates and oxygen are the products of photosynthesis. The entire process takes place in the presence of sunlight energy.

The carbon dioxide is the chief component for this reaction and the components of carbon are produced in the form of carbohydrates. The carbohydrates are the source of energy that are utilized by the producers for cellular metabolism and other functions. These get stored in the form of storage molecules in producers. Thus to increase the no. of storage molecules that producers can make Jaime need to increase the concentration of carbon dioxide in the ecosystem.

Answer:

Acceleration= final velocity-initial velocity/time taken

so

a= 10-2/10

a=-8/10

a= -.8 meters per second

Answer: 8.0 = 8 meters/seconds.

Explanation:

Started at= 10.0 meters/seconds.

Slows down= 2.0

10-2=8

Sandra is currently riding her bicycle at 8.0 meters/seconds.

*Mark brainlist* please :))

Answer:

1.43 Ω

Explanation:

Given that

Cross sectional area of the cord, A = 4.2*10^-7 m²

Distance meant to be used, L = 35 m

Resistivity of copper, p = 1.72*10^-8 Ωm

Temperature of copper, t = 20° C

Temperature coefficient of copper, t' = 0.004041 °C^-1

To solve this, we would use the resistance and resistivity formula

R = pL / A, on substituting

R = (1.72*10^-8 * 35) / 4.2*10^-7

R = 6.02*10^-7 / 4.2*10^-7

R = 1.43 ohm

Therefore, the resistance of the extension cord is 1.43 ohm

Answer:

v (minimum speed) = 2.90 m/sec.

[tex]\\ \\ maximum speed (v)= 6.57 m/sec.\\[/tex]

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a)  What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

[tex]m*g + T = m*v^2/R[/tex]

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

[tex]v = \sqrt{g*R}[/tex]

[tex]v = \sqrt{9.81*0.86}[/tex]

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the  force balance at bottom of circle is represented by the illustration:

[tex]T = m*g + m*v^2/R[/tex]

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass  m = 0.75 kg

∴

[tex]45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\[/tex]

c)

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because  at the lowest point; the tension in string will be maximum.

Answer:

1.57×10^-3 m^3/s

Explanation:

Please see the attached filw for a detailed and step by step solution of the given problem.

The attached file explicitly solves it.

Answer:

a) The radiation pressure on the Earth is 6.065x10⁻⁶Pa

b) The atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.

Explanation:

Given:

I = intensity of solar radiation = 1368 W/m²

Earth reflects 33%, therefore Earth absorbs 67%

P = pressure = 101 kPa = 1.01x10⁵Pa

c = speed of light = 3x10⁸m/s

Questions:

a) Find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead, P = ?

b) State how this quantity compares with normal atmospheric pressure at the Earth's surface

a) To solve, it is necessary to calculate the pressure exerted by both the reflected light and the light that is absorbed, in this way:

The pressure exerted by the reflected light:

[tex]P_{1} =\frac{2*0.33*I}{c} =\frac{2*0.33*1368}{3x10^{8} } =3.01x10^{-6} Pa[/tex]

The pressure exerted by the absorbed light:

[tex]P_{2} =\frac{0.67*I}{c} =\frac{0.67*1368}{3x10^{8} } =3.055x10^{-6} Pa[/tex]

The radiation pressure on the Earth:

Pt = P₁ + P₂ = 3.01x10⁻⁶ + 3.055x10⁻⁶ = 6.065x10⁻⁶Pa

b) Comparing with normal atmospheric pressure

[tex]Ratio=\frac{P_{atm} }{P_{t} } =\frac{1.01x10^{5} }{6.065x10^{-6} } =1.665x10^{10}[/tex]

According to this result, the atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.

Answer:

58.37 V

Explanation:

Given that

Number of winding on coil, N = 80 turns

Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m

Magnitude of magnetic force, B = 1.1 T

Time of rotation, t = 0.06 s

See the attachment for calculations

Answer:

ℰ[tex]Ф_{av}[/tex]=58.37V

Explanation:

Given:

Number of winding on coil 'N' = 80 turns

Area 'A' = 25 cm x 40 cm = 0.25 m x 0.4 m =>0.10m²

Magnitude of magnetic force 'B' = 1.1 T

Time ' t' = 0.06 s

The average magnitude of the induced emf is given by:

ℰ[tex]Ф_{av}[/tex]=N|ΔФB/Δt| => N |Ф[tex]Ф_{B,f}[/tex] - Ф[tex]Ф_{B,i}[/tex]| /Δt

The flux through the coil is Ф[tex]Ф_{B[/tex] = BA cos∅

ℰ[tex]Ф_{av}[/tex]=NBA |cos(∅[tex]Ф_{f}[/tex]) - cos(∅[tex]Ф_{i}[/tex])| /Δt

As the initial angle is ∅= 97-37 =>53° and the final angle is ∅=0°

ℰ[tex]Ф_{av}[/tex]=[tex]\frac{(80)(1.1)(0.1)|cos(0)-cos(53)|}{0.06}[/tex]

ℰ[tex]Ф_{av}[/tex]= [tex]\frac{(8.8)|1-0.002|}{0.06}[/tex]

ℰ[tex]Ф_{av}[/tex]=58.37V

Answer:[tex]\Delta L=0.0101\ m[/tex]

Explanation:

Given

Length of track [tex]L_o=28\ m[/tex] when

[tex]T_o=2^{\circ}C[/tex]

Coefficient of linear expansion [tex]\alpha =11\times 10^{-6}\ ^{\circ}C^{-1}[/tex]

When Temperature rises to [tex]T=35^{\circ}C[/tex]

[tex]\Delta T=35-2=33^{\circ}C[/tex]

and we know length expand on increasing temperature

[tex]L=L_o[1+\alpha \Delta T][/tex]

[tex]L-L_o=L_o\alpha \Delta T[/tex]

[tex]\Delta L=28\times 11\times 10^{-6}\times (33)[/tex]

[tex]\Delta L=0.0101=10.164\ mm[/tex]

(b)When rails are clamped thermal stress induced

we know [tex]E=\frac{stress}{strain}[/tex]

[tex]Stress=E\times strain[/tex]

[tex]Stress=20\times 10^{10}\times \frac{\Delta L}{L_o}[/tex]

[tex]Stress=20\times 10^{10}\times \frac{0.0101}{28}[/tex]

[tex]Stress=72.14\ MPa[/tex]

[tex]Stress=72.14\times 10^{6}\ N/m^2[/tex]

Answer:gamma rays

Explanation:

Out of all these spectrum listed in The question , gamma rays is closest to x-ray in The electromagnetic spectrum

Answer:

d) 0.011%

Explanation:

The probability for tunneling the barrier is given by the following formula:

[tex]P=exp(-2d\sqrt{\frac{2m_e(U_o-E)}{\hbar ^2} }\ )[/tex]      ( 1 )

me: mass of the electron

Uo: energy of the barrier

E: energy of the electron

d: thickness of the barrier

By replacing the values of the parameters in (1), you obtain:

[tex]P=exp(-2(0.20*10^{-9}m)\sqrt{\frac{2(9.11*10^{-31}kg)(100eV-80eV)(1.60*10^{-19}J)}{(1.055*10^{-34}Js)^2}})\\\\P=e^{-9.15}=1.08*10^{-4}\approx0.011\%[/tex]

hence, the probability is 0.011% (answer d)

Answer:

The distance travelled is [tex]S = 237.36 \ m[/tex]

Explanation:

From the question we are told that

    The distance of the airplane from the ground is  [tex]L = 24.0\ km[/tex]

     The angle subtended by the moon is [tex]\theta = 9.89*10^{-3} \ radians[/tex]

The distance traveled can be mathematically represented as

            [tex]S = R \theta[/tex]

Substituting values

            [tex]S = 24 *10^3 * 9.89*10^{-3}[/tex]

           [tex]S = 237.36 \ m[/tex]

Answer:

Explanation:

y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]

The angular velocity ω = 742 rad /s

wave function k = 6.98 rad /m

Amplitude A = 2.3 mm

y(x,t) = A cos( ωt + kx )

equation of wave reflected wave

y(x,t) = A cos( ωt - kx )

resultant standing wave

= y₁ +y₂

= A cos( ωt + kx ) +A cos( ωt - kx )

y(x,t) = 2 A cosωt cos kx

= 2 x 2.3 cos 742t .cos6.98x

= 4.6 mm . cos 742t .cos6.98x

The number of protons in the nucleus of an atom determines the element that the atom belongs to.

The subject of this question is Chemistry. The number of protons in the nucleus of an atom determines the species or element to which the atom belongs. This is because each element has a unique number of protons, known as the atomic number. The other options are not correct because:

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Answer:

1.27

Explanation:

Period of a pendulum T is

T = 2¶(l/g)^0.5

Where g is acceleration due to gravity

l is lenght of pendulum

For earth, T = 2.243 s

2.243 = 2 x 3.142 x (l/9.81)^0.5

0.36 = (l^0.5)/3.13

1.13 = l^0.5

l = 1.28 m

For planet X of the same lenght

Period T = 2.00 s

2 = 2 x 3.142 (1.28/gx)^0.5

0.32 = 1.13 / gx^0.5

3.53 = gx^0.5

gx^0.5 = 3.53

gx = 12.46 m/s^2

gx/gearth = 12.46/9.81 = 1.27

The ratio of the gravitational acceleration on planet X to that on Earth is 1.26, calculated using the periods of a pendulum on both planets and the relationship between period and gravitational acceleration.

The period of a simple pendulum is given by the formula T = 2*pi*sqrt(L/g) where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity. To find the ratio gX/gEarth, we can use the formula T_earth/T_x = sqrt(g_x/g_earth) (let's denote T_earth as the period on earth and T_x as the period on planet X).

Since we know the periods, we can substitute these values into the formula: 2.243s/2.000s = sqrt(g_x/g_earth), which gives us 1.1215 = sqrt(g_x/g_earth). To find the ratio g_x/g_earth, square both sides, giving (1.1215)² = g_x/g_earth. Thus, g_x/g_earth = 1.26.

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Answer:

The mass of the star would be  [tex]M = 2.644*10^{24} \ kg[/tex]    

Explanation:

From the question we are told that

     The angular speed is  [tex]w = 0.83\ rev/s = 0.83 * 2 \pi = 1.66 rad/s[/tex]

      The radius of the star is  [tex]r = 40km = 40 *1000 = 40 * 10^{3} m[/tex]    

Generally the minimum mass of the start is mathematically evaluated as

             [tex]M = \frac{r^3 w^2}{G}[/tex]

Where is the gravitational constant with a values of  [tex]G = 6.67*10^{-11} N \cdot m^2 /kg[/tex]

             [tex]M = \frac{(40*10^3)^3 * 1.66^2}{6.67*10^{-11}}[/tex]

             [tex]M = 2.644*10^{24} \ kg[/tex]    

Answer:

B. The transfer of energy from a hot object to a cold object

Explanation:

Answer:

The answer is option B

Explanation: