A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the ball (about 0.15 s). If the football has a mass of 0.44 kg, what average force does the punter exert on the ball?

Answer:

The lone pair of electrons occupy more space because the electrostatic force becomes weaker.

Explanation:

When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.

In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.    

A lone pair of electrons occupies more space around the central atom than bonding pairs due to greater electrostatic repulsions, which are not mitigated by the sharing of electrons with another atom. This concept also applies to multiple bonds, with higher electron density leading to increased space occupancy compared to single bonds and influencing molecular geometry.

The reason why a lone pair of electrons occupies more space around the central atom than a bonding pair of electrons relates to electrostatic repulsions. Lone pairs are not shared with another atom, therefore they tend to occupy a larger region of space due to increased repulsions of their negatively charged electrons. As illustrated in the case of a molecule like ammonia (NH3), the lone pair of electrons on the nitrogen atom occupies more space than the bonding pairs.

Additionally, this concept extends to multiple bonds such as double or triple bonds, which occupy more space around a central atom than a single bond. The higher electron density in multiple bonds leads to greater repulsions, which can influence the bond angles in a molecule, causing deviations from ideal geometry.

The order of space occupied from largest to smallest is as follows: lone pair > triple bond > double bond > single bond. This hierarchy is essential for understanding molecular shape and bond angles, and is exemplified by deviations in expected bond angles due to these repulsions.

Answer:

1800/300 = 6ropes

Explanation:

The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.

Answer:

a. [tex]6.032\times10^{-6}C/m^2[/tex]

b.[tex]6.816\times10^5N/C[/tex]

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as [tex]\sigma[/tex]

[tex]\sigma=\frac{Q}{4\pi r_{out}^2}[/tex], and the strength of the electric field outside the sphere [tex]E=\frac{\sigma _{new}}{\epsilon _o}[/tex]

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

[tex]\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2[/tex]  #surface charge outside sphere.

[tex]\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2[/tex]

Hence, the new charge density on the outside of the sphere is [tex]6.032\times10^{-6}C/m^2[/tex]

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be [tex]6.032\times10^{-6}C/m^2[/tex],

[tex]E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C[/tex]

Hence, the strength of the electric field just outside the sphere is [tex]6.816\times10^5N/C[/tex]

Answer:

[tex]W=K_f-K_i[/tex]

Explanation:

The work done on a particle by external forces is defined as:

[tex]W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,[/tex]

According to Newton's second law [tex]F=ma[/tex]. Thus:

[tex]W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\[/tex]

Acceleration is defined as the derivative of the speed with respect to time:

[tex]W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,[/tex]

Speed is defined as the derivative of the position with respect to time:

[tex]W=m\int\limits^{v_f}_{v_i} v \cdot dv \,[/tex]

Kinetic energy is defined as [tex]K=\frac{mv^2}{2}[/tex]:

[tex]W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i[/tex]

The skydiver has more force acting downward due to gravity than upward from air resistance, and thus he's accelerating downwards. You calculate the acceleration by subtracting the air resistance from the skydiver's weight and dividing by his mass, in this case, yielding an acceleration of 7.36 m/s².

The skydiver's situation can be understood by taking into account the forces acting on him, mainly his weight and the air resistance. The weight of the skydiver is determined by multiplying his mass by the gravity constant, 9.81 m/s². This gives a weight of 101 kg × 9.81 m/s² = 991.81 N. The skydiver is said to be experiencing one-fourth that amount in upward force due to air resistance, which would be approximately 247.95 N.

In situations where opposing forces are equal, an object doesn't accelerate and we would expect the skydiver to be in a state of terminal velocity. However, in this case, there's more force acting downward (gravity) than upward (air resistance), so the skydiver is accelerating downwards. To find the amount of that acceleration, we subtract the air resistance from the skydiver's weight, then divide that by his mass. Therefore, (991.81 N - 247.95 N) / 101 kg = 7.36 m/s².

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Answer:

11.714 kW

Explanation:

Here is the complete question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?

Solution

Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where  mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration

F = m(a + gsinθ)  

When the car is moving at constant speed, a = 0

So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N

Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW

The force needed to pull a 950 kg ore car up a 34.0° inclined mine shaft is approximately 5518.493 N, involving both the acceleration force and the component of gravitational force parallel to the incline.

The question is about calculating the acceleration and force needed to pull a loaded ore car up a mine shaft inclined at 34.0° above the horizontal.

Step-by-Step Explanation:

Therefore, the total force needed to pull the ore car up the inclined shaft is approximately 5518.493 N.

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m

Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus(Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F/(AY)

So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)

When expressed in percentage, it becomes 0.00729%

An accurate calculation for the percent decrease in the stool legs' length under a 70 kg man's weight requires knowledge of the materials properties and dimensions of the stool. General concepts of stress, strain, and deformation in materials are discussed, including their relationship to applied forces and material properties as described by Hooke's Law.

To determine the percent decrease in the length of the stool legs when a 70 kg man sits on it, we need additional information such as the material properties of the stool legs and any relevant physical dimensions. However, with the given scenario, we can talk generally about stress and strain in materials and how they are calculated in relation to force and deformation. Stress is the force per unit area applied to a material, while strain is the deformation experienced by the material relative to its original length. For an object like a stool leg, the deformation (change in length) due to a person sitting on it could be calculated using Hooke's Law if the deformation remains within the linear elastic range of the material.

Without specific data on the stool's material, cross-sectional area, and elastic modulus, we cannot calculate the exact percent decrease in length of the stool's legs. If such information were provided, we could use the formula σ = F/A, where σ is the stress, F is the force, and A is the cross-sectional area; and the formula ε = δL/L, where ε is the strain, δL is the change in length, and L is the original length. Combined with Hooke's Law (σ = Eε, E being the elastic modulus), we could find the deformation and thus the percent decrease in length.

Answer:

128 N

Explanation:

Using

F = Gm'm/r²....................... Equation 1

Where F = Force, G = Universal constant, m = mass of the human, m' = mass of the moon, r = radius of the moon

Given: m = 75 kg, m' = 7.4×10²² kg, r = 1.7×10⁶ m

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 1

F =  (6.67×10⁻¹¹ )(75)(7.4×10²²)/(1.7×10⁶)²

F = (3.7×10¹⁴)/(2.89×10¹²)

F = 1.28×10²

F = 128 N

Explanation:

Below is an attachment containing the solution.

Answer:

a) 44.8J

b) 44.8J

c) 127cm

Explanation:

a) The elastic potential energy of a compressed spring is given by the formula:

[tex]U_e=\frac{1}{2} kx^{2}[/tex]

Where U_e is the elastic potential energy, k is the spring constant and x is the distance the spring is compressed. In this case, we have:

[tex]U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm[/tex]

To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:

[tex]U_e=4480Ncm=4480N(0.01m)=44.8J[/tex]

In words, the elastic potential energy of the compressed spring is 44.8J.

b) Using the law of conservation of mechanical energy, we have that:

[tex]E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f[/tex]

Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that [tex]U_{g0}=0;K_0=0;U_{ef}=0;K_f=0[/tex] because in t=0 the block has no speed and is in tis lowest point; and in t=t_f the block has stopped and isn't in contact with the spring. So, our equation is reduced to:

[tex]U_{gf}=U_{e0}\\\\U_{gf}=44.8J[/tex]

So, the gravitational potential energy of the block in its highest point is 44.8J.

c) Using the gravitational potential energy formula, we have:

[tex]U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm[/tex]

Using trigonometry, we can compute the distance between the release point and its highest point:

[tex]d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm[/tex]

Answer:

      [tex]2.42liter[/tex]

Explanation:

The question is incomplete; you need some additional data.

I will assume the missing data from a similar question and keep the final pressure of 703 torr given in your question.

An ideal gas at a pressure of 1.20 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.720 L as shown here. When the stopcock is opened the gas expands into the empty bulb

If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas?

Since the amount of gas and the temperature remain constant, we may use Boyle's law:

             [tex]PV=constant\\\\P_1V_1=P_2V_2[/tex]

Form the data:

        [tex]P_1=1.20atm\\\\P_2=703torr\\\\V_1=unknown\\\\V_2=V_1+0.720liter[/tex]

1. Convert 703 torr to atm:

   [tex]P_2=703torr\times 1atm/760torr=0.925atm[/tex]

2. Sustitute the data in the equation:

       [tex]1.2atm\times V_1=0.925atm\times(V_1+0.720liter)[/tex]

3. Solve:

      [tex]1.2V_1=0.925V_1+ 0.666liter\\\\0.275V_1=0.666liter\\\\V_1=2.4218liter\approx 2.42liter[/tex]

The answer is rounded to 3 significant figures, according to the data.

The maximum potential difference between points A and B is 100 V, and the maximum energy stored in the three-capacitor arrangement is determined by the capacitor with the smallest capacitance.

The maximum potential difference (voltage) between points A and B in a series combination of capacitors is limited by the capacitor with the lowest breakdown voltage. Given C2 = 20.0 μF and C3 = 25.0 μF, and no capacitor can exceed 100 V, the maximum potential difference that can exist between points A and B is 100 V.

The maximum energy that can be stored in the three-capacitor arrangement is determined by the capacitor with the least energy storage capability. In this case, the capacitors are in series, so the total energy stored in the arrangement is the energy stored by the capacitor with the smallest capacitance, which is 20 μF, at the maximum allowed voltage of 100 V.

Answer:

the membrane potential

Explanation:

typical values of membrane potential are in the range -40 mv to-70 mv

Answer:

t = 0.1 s

Explanation:

Given:

- The distance crushed (Stopping-Distance) s = 1.20 m

- The mass of the car m = 1,550 kg

-  The initial velocity vi = 24.0 m/s

Find:

- How long does the collision last t?

Solution:

- The stopping distance s is the average velocity v times time t as follows:

                                s = t*( vf + vi ) / 2

Where,

             vf = 0 m/s ( Stopped )

                                s = t*( vi ) / 2

                                t = 2*s / vi

                                t = 2*1.20 / 24

                                t = 0.1 s

Answer:

t=0.1seconds.

Explanation:

The mass of the car m = 1,550 kg  

We know the stopping distance, 1.20m,

we know the final velocity, 0m/s (its stopped),

the starting velocity, 24m/s.

d=t(v2+v1)/2

Rearange to solve for t, and remove the v2^2 as its zero

t=2d/v1  

t=2(1.20m)/(24m/s)

t=0.1seconds.

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between  loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

[tex]AC^2=AB^2+BC^2[/tex]

[tex]AC^2=2.00^2+5.50^2[/tex]

[tex]AC=\sqrt{(2.00^2+5.50^2)}[/tex]

[tex]AC=5.85\ m[/tex]

We need to calculate the path difference

Using formula of path difference

[tex]\Delta x=AC-BC[/tex]

Put the value into the formula

[tex]\Delta x=5.85-5.50[/tex]

[tex]\Delta x=0.35\ m[/tex]

We need to calculate the lowest possible frequency of sound

Using formula of frequency

[tex]f=\dfrac{nv}{\Delta x}[/tex]

Put the value into the formula

[tex]f=\dfrac{1\times340}{0.35}[/tex]

[tex]f=971.4\ Hz[/tex]

Hence, The lowest possible frequency of sound is 971.4 Hz.

A

Explanation:

fred ball hit the pin the time for all just put a hit in his right knee and do not need to be on bed at the carbon footprint in his first year of practice at all but his teammates were also on bed and you were a good friend to be with you cute guys that he is the most likely and do you have the ability for you I don't want him on you and

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit. The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit. The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.

The time it takes for Betty's ball to hit the pins is equal to that for Barney's ball to hit.

The time it takes for Fred's ball to hit the pins is equal to that for Wilma's ball to hit.

The time it takes for Wilma's ball to hit the pins is greater than that for Betty's ball to hit.

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Explanation:

The statement " Close spacing represents greater current densities" is true about the spacing of the streamlines in a wire (option F)

Why is this correct?

In a scenario where a conductor is wider on the left and narrower on the right, the electric field lines and current density are depicted by streamlines. With current flowing from the wider end to the narrower end, the total charge and current remain constant. However, the current density fluctuates, being higher at the narrower end.

Hence, when observing streamlines, closer spacing within the wire signifies a higher current density, specifically in the narrower sections compared to the wider ones.

Complete question:

Which is true about the spacing of the streamlines in a wire?

A. Wide spacing represents faster random-motion velocities.

B. Wide spacing represents greater electric field vectors.

C. Wide spacing represents greater current densities.

D. Close spacing represents faster random-motion velocities.

E. Close spacing represents greater electric field vectors.

F. Close spacing represents greater current densities.

Answer: FALSE

Explanation: Could you help me with a question?

Answer:

The frequency 400 hz is not possible .

Explanation:

Given that,

Frequency = 200 hz

Length = l

Suppose, The given frequencies are,

600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.

The possible resonance frequencies are

We need to calculate the fundamental frequency

Using formula of fundamental frequency for pipe

[tex]F=\dfrac{nv}{4L}[/tex]

Where, n = odd number

Put the value of frequency

[tex]200= \dfrac{nv}{4L}[/tex]

We need to calculate the first over tone

Using formula of fundamental frequency

n = 3,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{2}=3\times\dfrac{v}{4l}[/tex]

[tex]F_{2}=3\times200[/tex]

[tex]F_{2}=600\ Hz[/tex]

We need to calculate the second over tone

Using formula of fundamental frequency

n = 5,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{3}=5\times200[/tex]

[tex]F_{3}=1000\ Hz[/tex]

We need to calculate the third over tone

Using formula of fundamental frequency

n = 7,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{4}=7\times200[/tex]

[tex]F_{4}=1400\ Hz[/tex]

We need to calculate the fourth over tone

Using formula of fundamental frequency

n = 9,

[tex]F=\dfrac{nv}{4L}[/tex]

Put the value into the formula

[tex]F_{5}=9\times200[/tex]

[tex]F_{5}=1800\ Hz[/tex]

Hence, The frequency 400 hz is not possible .

Final answer:

In a pipe that is closed at one end and open at the other, only odd multiples of the fundamental frequency will resonate. Frequencies that are even multiples of the fundamental 200 Hz, such as 400 Hz, 600 Hz, 800 Hz, will not resonate.

Explanation:

The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 Hz. This pipe supports harmonic frequencies following the sequence fn = n(v/4L), where n = 1, 3, 5..., v is the speed of sound, and L is the length of the pipe. Therefore, frequencies that will not resonate in the same pipe are those that are even multiples of the fundamental frequency, such as 400 Hz, 600 Hz, 800 Hz, and so on.

Answer: into the atmosphere

Explanation:

As this energy is released into the atmosphere, bubbles of warm air is formed which is released and it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere, and is known as....

convection

While convection is a form of heat transfer from one place to the other involving the movement of fluids. So in the case narrated above the fluid which serves as a medium is Air.

Answer:

Alfred Wegener

Explanation:

Alfred Wegener was a German scientist who first discovered the idea of continental drift.

The continental drift hypothesis says that the large continents drift from one location to another over the broad ocean water bodies with respect to the fixed poles. This was the first step in understanding the interior of the earth and how the tectonic activities take place on earth.

He contributed many pieces of evidence in order to support this hypothesis, but it was initially not accepted as he was not able to explain the main mechanism for the continental motion.

Answer:

0.27J

Explanation:

[tex]C_eq= C_1 + C_2\\= 2+4\\= 6UF\\U = (1/2) CV^2\\= (1/2)(6 - 6)(300 * 300)\\= 0.27J[/tex]

Answer:

The ride is above 22m in height for 1.33 minutes.

Explanation:

Let's first find the height required above the boarding platform for the ride to be 22 m above the ground:

Height required = 22 - 5 = 17 m

We can now, using a right angled triangle of height equal to the Ferris wheel radius, calculate the angle from the vertical axis to achieve this height:

Height of triangle = 15 - (17 - 15) = 13 m

Hypotenuse of triangle = radius = 15 m

Angle from the vertical:

Cos( Angle ) = base / hypotenuse = 13 / 15

Angle = 29.92 °

Multiplying this angle by 2 we get the total angle through which the ride is at the required height:

Total Angle = 29.92 * 2 = 59.85 °

To take out the time we can now simply multiply the ratio of this angle /360 by the time taken for one complete revolution:

Time = [tex]\frac{59.85}{360} * 8[/tex]

Time = 1.33 minutes

Explanation:

Answer:

The force that would be applied on the rope just to start moving the wagon is 122 N

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

          μ is the coefficient of friction ( μs, in this case, static friction);

          m  is mass of the object and;

          g is the acceleration due to gravity( a constant equal to 9.81 m/[tex]s^{2}[/tex])

from the equation we are provide with;

       μs  = 0.25

       m = 50 kg

       g =  9.81 m/[tex]s^{2}[/tex]

      F =?

Using equation 1

F = 0.25 x 50 kg x  9.81 m/[tex]s^{2}[/tex]

F = 122.63 N  

Therefore a force of 122 N must be applied to the rope just to start the wagon.

Explanation:

Below is an attachment containing the solution.

Answer:

THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.

Explanation:

Water flows more rapidly out of the ground-floor faucet.

Learn more about municipal water supply here:

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Explanation:

    F = k×[tex]\frac{q1q2}{r2}[/tex]

where, K is coulombs constant, which is equal to -                                  9 x10^9 [tex]Nm^2/C^2.[/tex]

Answer:

They have a frequency of 2 hertz

Explanation:

Frequency is the number of occurrences of a repeating event per unit of time and Wavelength is the distance from one crest to another, or from one trough to another.

In the question, it is stated that the leaf does two full up and down bobs, this means that it completes 2 full cycles in one second. Therefore, its frequency is 2/s

where, s⁻¹ is hertz

so, They have a frequency of 2 hertz

Answer: A = square root (2/L)

Explanation: find the attached file for explanation

Final answer:

The wave function of a particle in a one-dimensional box needs to be normalized. The normalization condition requires that the value of A for the wave function Ψ(x) = Asin(πx/L) is found to be A = √(2/L) after solving the normalization integral.

Explanation:

The wave function Ψ(x) of a particle in a one-dimensional box of width L must be normalized so that the total probability of finding the particle within the box is 1. This normalization condition implies that the integral of the square of the absolute value of the wave function over the interval from 0 to L should equal 1.

The normalization integral for the given wave function Ψ(x) = Asin(πx/L) is:

∫ |Asin(πx/L)|² dx = A² ∫ sin²(πx/L) dx = 1

When you solve the integral from 0 to L, the result is:

A² * L/2 = 1

Therefore, solving for A, we get:

A = √(2/L)

So the value of A in terms of L is √(2/L).

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: [tex]F = 49.06N[/tex]

Pebble: [tex]F = 29.44N[/tex]

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: [tex]a =9.8m/s^{2}[/tex]

Pebble:  [tex]a =9.8m/s^{2}[/tex]

Explanation:

The universal law of gravitation is defined as:

[tex]F = G\frac{m1m2}{r^{2}}[/tex]  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

Case for the rock [tex]m = 5.0 Kg[/tex]:

m1 will be equal to the mass of the Earth [tex]m1 = 5.972×10^{24} Kg[/tex] and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth [tex]r = 6371000m[/tex].

[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}[/tex]

[tex]F = 49.06N[/tex]

Newton's second law can be used to know the acceleration.

[tex]F = ma[/tex]

[tex]a =\frac{F}{m}[/tex] (2)

[tex]a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}[/tex]

[tex]a =9.8m/s^{2}[/tex]

Case for the pebble [tex]m = 3.0 Kg[/tex]:

[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}[/tex]

[tex]F = 29.44N[/tex]

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

[tex]a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}[/tex]

[tex]a =9.8m/s^{2}[/tex]

(a) The magnitude of the gravitational force on the rock is 49 N, while the magnitude of the gravitational force on the pebble is 2.94 × 10^-3 N. (b) The magnitude of the acceleration for both objects when released is 9.8 m/s^2.

(a) Gravitational Force:

The magnitude of the gravitational force exerted on an object near the surface of the Earth can be calculated using the formula:

F = mg

where F is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

For the rock, m = 5.0 kg:

F = mg = (5.0 kg) * (9.8 m/s^2) = 49 N

For the pebble, m = 3.0 × 10-4 kg:

F = mg = (3.0 × 10-4 kg) * (9.8 m/s^2) = 2.94 × 10-3 N

(b) Acceleration:

The magnitude of the acceleration of an object when released can be calculated using Newton's second law:

F = ma

Rearranging the equation to solve for acceleration, we have:

a = F/m

Substituting the values we calculated in part (a) for the gravitational force exerted on each object, we can determine the magnitude of acceleration.

For the rock, F = 49 N and m = 5.0 kg:

a = (49 N) / (5.0 kg) = 9.8 m/s^2

For the pebble, F = 2.94 × 10-3 N and m = 3.0 × 10-4 kg:

a = (2.94 × 10-3 N) / (3.0 × 10-4 kg) = 9.8 m/s^2

Learn more about Gravitational force and acceleration here:

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