What fraction of the copper's electrons has been removed? Each copper atom has 29 protons, and copper has an atomic mass of 63.5.

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

[tex]dtan13^0 = 0.231d[/tex]

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

[tex]dtan4^0 = 0.07d [/tex]

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

To find the distance to the 200-foot tall monument from a specified point, we use tangent trigonometric functions with the given angles of elevation and depression. The total horizontal distance is calculated by separately finding the distances to the top and the bottom of the monument and combining them.

The question involves solving a problem using trigonometry to find the distance to a monument given the angles of elevation and depression from a certain point. Since the height of the monument is given as 200 feet, and we have the angles of elevation (13°) to the top of the monument and the angle of depression (4°) to the bottom, we can use trigonometric functions to calculate the distances to the top and the bottom of the monument separately and then sum them to find the total distance from the observer to the monument.

To find the horizontal distances (D) from the observer to the top and bottom of the monument, we can use the tangent function (tan) from trigonometry. The tangent of the angle of elevation (13°) is equal to the height of the monument divided by the distance to the top (Dtop), and the tangent of the angle of depression (4°) is equal to the height from the window to the base of the monument divided by the distance to the bottom (Dbottom).

Assuming the window is at the same height as the base of the monument (which is not given explicitly, but implied by the angle of depression), we get two equations:

We can then solve for Dtop and Dbottom using these equations and add the two distances to find the total horizontal distance to the monument.

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Answer:

v= 14.22 m/s

Explanation:

Given that

u = 12.1 m/s

a=0.35   m/s².

t= 6.07 s

We know v = u +at

v=final velocity

u=initial velocity

t=time

a=acceleration

Now by putting the values in the above equation

v= 12.1 + 0.35 x 6.07 m/s

v= 14.22 m/s

Therefore the final velocity will be 14.22 m/s.

The final velocity of a bicycle racer with an initial velocity of 12.1 m/s and an acceleration of 0.350 m/s² over 6.07 s is approximately 14.22 m/s.

To compute the final velocity of a bicycle racer sprinting towards the end of a race, we can employ the kinematic equation for velocity under constant acceleration:

v = u + at

Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Given that the racer has an initial velocity u of 12.1 m/s, accelerates at a rate a of 0.350 m/s² for a duration t of 6.07 s, we substitute these values into the equation:

v = 12.1 m/s + (0.350 m/s² × 6.07 s)

Calculation:

v = 12.1 m/s + (0.350 m/s² × 6.07 s)
v = 12.1 m/s + 2.1245 m/s
v = 14.2245 m/s

The final velocity of the racer is approximately 14.22 m/s.

Answer:

[tex]a=-3449.67\frac{m}{s^2}[/tex]

Explanation:

The car is under an uniforly accelerated motion. So, we use the kinematic equations. We calculate the acceleration from the following equation:

[tex]v_f^2=v_0^2+2ax[/tex]

We convert the initial speed to [tex]\frac{m}{s}[/tex]

[tex]39\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=10.83\frac{m}{s}[/tex]

The car stops, so its final speed is zero. Solving for a:

[tex]a=\frac{v_0^2}{2x}\\a=-\frac{(10.83\frac{m}{s})^2}{2(1.7*10^{-2}m)}\\a=-3449.67\frac{m}{s^2}[/tex]

To calculate the acceleration of the car, use the formula a = (vf - vi) / t. Convert the initial velocity and distance to the appropriate units before substituting them into the formula.

To calculate the acceleration of the car, we can use the formula:

a = (vf - vi) / t

Where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time taken. In this case, the initial velocity is 39 km/h, which is converted to m/s by vi = 39 km/h * (1000 m/1 km) * (1 h/3600 s). The final velocity is 0 m/s since the car stops. The time taken can be found by t = d / vi, where d is the distance and vi is the initial velocity. The distance is given as 1.7 cm, which is converted to m by d = 1.7 cm * (1 m/100 cm). Substituting these values into the formula, we get the acceleration of the car.

So the acceleration of the car is approximately -8.45 m/s^2.

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Explanation:

For a pipe with one end open ,we have the formula for fundamental frequency as

[tex]f_1= \frac{v}{4L}[/tex]

v= velocity of sound in air =340 m/s

L= length of pipe

hence, [tex]L= \frac{v}{4f_1}[/tex]

given f_1 = 440 Hz

Substituting , L = 340/(4×440) = 0.193 m

a)  Let the bottle be filled to height , h.

Given that height of bottle ,H = 25 cm = 0.25 m

Also we found out that length of pipe L = 0.193 m

So, we have   h = H - L

= 0.25 - 0.193 = 0.057 m = 5.7 cm.

Answer:

(a). The bottle filled at the height is 5.5 cm

(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.

Explanation:

Given that,

Frequency = 440.0 Hz

Height of bottle = 25.0 cm

Suppose,

Let the bottle be filled to height h.

For a pipe with one end open,

We need to calculate the length of the pipe

Using formula of fundamental frequency

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

[tex]L=\dfrac{v}{4f}[/tex]

Where, L = length

v = speed of sound

Put the value into the formula

[tex]L=\dfrac{343}{4\times440}[/tex]

[tex]L=0.195\ m[/tex]

(a). We need to calculate the height

Using formula of height

[tex]h=H-l[/tex]

Where, H = height of bottle

l = length of pipe

Put the value into the formula

[tex]h=25.0\times10^{-2}-0.195[/tex]

[tex]h=0.055\ m[/tex]

[tex]h=5.5\ cm[/tex]

(b). We need to calculate the frequency of the nest higher harmonic of this bottle

Using formula of frequency

[tex]f_{next}=nf_{1}[/tex]

Where, [tex]f_{1}[/tex]=fundamental frequency

Put the value into the formula

[tex]f_{2}=2\times440[/tex]

[tex]f_{2}=880\ Hz[/tex]

Hence, (a). The bottle filled at the height is 5.5 cm

(b). The frequency of the nest higher harmonic of this bottle is 880 Hz.

Answer:

The electric potential at the center of the sphere is 36 V.

Explanation:

Given that,

Radius R= 0.600 m

Distance D = 1.20 m

Electric potential = 18.0 V

We need to calculate the electric potential

Using formula of electric potential

[tex]V=\dfrac{kq}{r}[/tex]

Put the value into the formula

[tex]18.0=\dfrac{9\times10^{9}\times q}{1.20}[/tex]

[tex]q=\dfrac{18.0\times1.20}{9\times10^{9}}[/tex]

[tex]q=2.4\times10^{-9}\ C[/tex]

We need to calculate the electric potential at the center of the sphere

Using formula of potential

[tex]V=\dfrac{kq}{r}[/tex]

Put the value into the formula

[tex]V=\dfrac{9\times10^{9}\times2.4\times10^{-9}}{0.600}[/tex]

[tex]V=36\ V[/tex]

Hence, The electric potential at the center of the sphere is 36 V.

The electric potential at the center of the sphere is 36 V.

The given parameters;

The magnitude of the charge is calculated as follows;

[tex]V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\q = \frac{18 \times 1.2}{9\times 10^9} \\\\q = 2.4 \times 10^{-9} \ C[/tex]

The electric potential at the center of the sphere;

[tex]V = \frac{kq}{r} \\\\V = \frac{9\times 10^9 \times 2.4 \times 10^{-9}}{0.6} \\\\V = 36 \ V[/tex]

Thus, the electric potential at the center of the sphere is 36 V.

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Answer:

Explanation:

It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.

If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.

If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.

If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere                        

A person inside a charged, hollow, metallic sphere, known as a Faraday cage, would not be harmed upon touching the interior, regardless of their own charge, due to the neutralization of the electric field inside the conductor.

A person placed inside a large, hollow, metallic sphere that is insulated from the ground and charged will not be harmed upon touching the inside of the sphere. The phenomenon that explains this is known as the Faraday cage effect. According to this principle, an external static electric field will cause the charges within a conductor to rearrange themselves in such a way that the electric field inside the conductor cancels out.

Now, if the person inside the sphere also has a charge of the opposite sign to that of the sphere, an interesting interaction occurs. However, due to the conductor's nature, the electric field inside the metallic sphere remains null. The charges within the conductor would still redistribute to neutralize the electric field within the conductor. Therefore, if a person inside touched the inner surface of the sphere, they would not be directly harmed by the electric field, as it is neutralized within the conductor. The charges from the person would likely redistribute on the sphere's inner surface to maintain electrostatic equilibrium without causing harm.

Answer:

elastic force and weight are related to the acceleration of the System.

Explanation:

The relationship between these two forces can be found with Newton's second law.

        [tex]F_{e}[/tex] - W = m a

        K x - m g = m a

We see that elastic force and weight are related to the acceleration of the System.

If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur

The relationship between the applied force of a hanging mass and the spring force of the spring is governed by Hooke's law, where the force is proportional to displacement. At equilibrium, the spring force balances the gravitational force, and any displacement results in harmonic oscillation.

The relationship between the applied force of a hanging mass on a spring and the spring force is described by Hooke's law. When a mass is attached to a vertical spring, it stretches until the spring force equals the force of gravity on the mass. This point is called the equilibrium position. The spring force at this point is k times the displacement from the spring's unstretched length, where k is the spring constant. If the mass is displaced from this equilibrium position, a restoring force acts to return the mass to equilibrium, leading to harmonic oscillation.

In the context of springs, two scenarios arise. When the spring is relaxed, there is no force on the block, indicating an equilibrium situation. Upon stretching or compressing the spring, Hooke's law predicts the force on the block as being proportional to the displacement from its equilibrium position, x, with force F being -Kx. Here, -K represents the negative spring constant, signifying a restoring force opposite to the direction of displacement.

For a massive continuous spring with negligible gravitational effect compared to the spring force (KL » mg), when the top of the spring is driven up and down, the position of the bottom can be described as a function of time. This shows that both vertical and horizontal spring-mass systems follow similar dynamics and obey the laws governing harmonic motion.

Answer:

The speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s

Explanation:

qV = 0.5Mv²

where;

V is the potential difference, measured in Volts

q is the charge of the electron in Coulomb's = 1.6 × 10⁻¹⁹ C

Mass is the mass of the electron in kg =  9 × 10⁻³¹ kg

v is the velocity of the electron in m/s

Applying coulomb's law, we determine the Potential difference V

V = kq/r

V = (8.99X10⁹ * 1.6 × 10⁻¹⁹)/(1X10⁻⁹)

V = 14.384 X 10¹⁹ V

The speed of the electron can be determined as follows;

v² = (2qV)/M

v = √(2qV)/M)

v = √(2*1.6 × 10⁻¹⁹* 14.384 X 10¹⁹)/(9 × 10⁻³¹)

v = √(5.1143 X 10³¹) = 7 X 10¹⁵m/s

Therefore, the speed of the third electron when it was far away from other electrons is 7 X 10¹⁵m/s

Answer: The question is incomplete as some other details are missing. Here is the complete question ; A helium atom falls in a vacuum . The mass of a helium atom is 6.64 x 10-27 kg . (a) From what height must it fall so that its translational kinetic energy at the bottom equals the average translational kinetic energy of a helium molecule at 300 K? (b) At what temperature would the average speed of helium atoms equal the escape speed from the Earth, 1.12 x 104 m/s.

a) Height = 95.336Km

b) Temperature = 20118.87K

Explanation:

The detailed steps and calculations is shown in the attached file.

The helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.

It is the required energy to accelerate the particle or object from the rest to achieve the given velocity.

The height can be calculated by the formula,

[tex]h = \dfrac 23 \times \dfrac {kt}{mg}[/tex]

Where

h - height =?

k - constant =[tex]1.38\times 10^{-23}[/tex] J/k

t - temperature = 300 k

m - mass

g - gravitational acceleration = 9.8 m/s^2

Put the values in the formula,

[tex]h = \dfrac 23 \times \dfrac {1.38\times 10^{-23}\times 300 }{6.64 \times 10^{-27}\times 9.8}\\\\h = 95335.4 {\rm\ m \ \ or}\\\\h = 95.336\rm \ km[/tex]

Therefore, the helium atom must fall from 95.336 km to achieve average translational kinetic energy at 300 K.

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Answer:

As potential energy depends on height

P.E=mgh

m and g remains same

just before release it has maximum height,thus option D it has maximum potential energy.

The ball has the most potential energy when it is at the top of the building D. before it is released.

Potential energy is stored energy that an object possesses due to its position, state, or condition. It can be gravitational, elastic, chemical, or electrical. The energy is converted to kinetic energy when the object moves or undergoes a change in its physical or chemical state.

The ball has the most potential energy when it is at the top of the building before it is released. Potential energy depends on an object's height and mass. When the ball is at the top of the building, it has the maximum height and therefore the most potential energy. As the ball falls, its potential energy decreases and is converted into kinetic energy.

Answer:

(a) WaveLength

Explanation:

Compared with ultraviolet radiation, infrared radiation has greater wavelength. Option A.

Compared with ultraviolet radiation, infrared radiation has greater wavelength.

Wavelength refers to the distance between two consecutive peaks or troughs of a wave. In the electromagnetic spectrum, infrared radiation has longer wavelengths than ultraviolet radiation.

Ultraviolet radiation has shorter wavelengths and higher energy, while infrared radiation has longer wavelengths and lower energy.

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Answer:

76.3 m

Explanation:

We are given that

Initial speed of the ball,u=(15+A)m/s

Height of cliff,h=(25.9+B) m

We have to find the distance from the base of the cliff the ball will land in the water below.

A=4 and B=54

Distance=[tex]u\sqrt{\frac{2h}{g}}[/tex]

Using the formula and substitute the values

[tex]D=(15+4)\sqrt{\frac{2(25+54)}{9.8}}[/tex]

Because [tex]g=9.8m/s^2[/tex]

[tex]D=19\sqrt{\frac{158}{9.8}}[/tex]

D=76.3

Hence, the distance from the base of the cliff the ball will land in the water below=76.3 m

Answer:

T

Explanation:

I learned this in 5th grade

It is true that an electron moves to a higher energy level in an atom after absorbing a photon of a specific energy.

The transfer of a bound electron to a more energetic, but still bound, state is known as electron excitation.

This can be accomplished through photoexcitation, in which the electron absorbs a photon and absorbs all of its energy, or through electrical excitation, in which the electron receives energy from another, more energetic electron.

An electron can become excited if it receives additional energy, such as when it absorbs a photon, or light packet, or collides with a nearby atom or particle.

Electrons can absorb energy (discretely) to move from lower to higher energy states.

When they reach higher energy states, they become unstable and emit photons to return to their ground state (of specific energy).

Thus, the given statement is true.

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Final answer:

The speed of the putty just before it strikes the ceiling is approximately 0.342 m/s, and it takes approximately 0.97 seconds to reach the ceiling.

Explanation:

To calculate the speed of the putty just before it strikes the ceiling, we can use the equation for vertical motion:

v = u + gt

Where:

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity (-9.8 m/s^2)

t is the time

In this case, the object is thrown upwards, so the final velocity when it reaches the ceiling will be zero. The initial velocity is 9.50 m/s and the acceleration due to gravity is -9.8 m/s^2. Substitute these values into the equation and solve for t:

0 = 9.50 - 9.8t

t = 0.97 seconds

Therefore, the speed of the putty just before it strikes the ceiling is 9.50 - 9.8(0.97) = -0.342 m/s. Since speed cannot be negative, the actual speed is 0.342 m/s.

To calculate the time it takes for the putty to reach the ceiling, we can use the same equation and solve for t:

0 = 9.50 - 9.8t

t = 0.97 seconds

So, it takes the putty approximately 0.97 seconds to reach the ceiling.

Final answer:

The speed of the putty just before it strikes the ceiling is 9.50 m/s, the same as its initial speed due to conservation of energy. The time it takes for the putty to reach the ceiling can be calculated using kinematic equations.

Explanation:

The problem presented concerns the motion of a glob of putty thrown upward and involves calculations based on the principles of kinematics in physics.

Answer to Part (a): What is the speed of the putty just before it strikes the ceiling?

The speed of the putty just before it strikes the ceiling can be found by using the kinematic equation for velocity under constant acceleration (gravity). Since the putty is thrown upwards, it will slow down under the influence of gravity until it reaches its maximum height. At this point, it will start to fall back down, accelerating under gravity until it hits the ceiling. Assuming no energy loss, the speed of the putty just before it strikes the ceiling will be the same as its initial speed when it leaves the hand, which is 9.50 m/s.

Answer to Part (b): How much time from when it leaves your hand does it take the putty to reach the ceiling?

The time it takes for the putty to reach the ceiling can be calculated using the kinematic equation s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration due to gravity (approximately -9.81 m/s² when upwards is considered positive), and t is the time. Solving for t gives us the time taken for the putty to reach the height of 3.60 m.

Answer:

(a). The shear stress on the pipe wall is 0.2062 lb/ft²

(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²

(c).  The shear stress at the distance 0.5 in away from the pipe wall is zero.

Explanation:

Given that,

Diameter = 1 in

Pressure = 0.55 psi

Length = 8 ft

We need to calculate the radius of the pipe

Using formula of radius

[tex]r=\dfrac{D}{2}[/tex]

Put the value into the formula

[tex]r=\dfrac{1}{2}[/tex]

[tex]r=0.5\ in[/tex]

(a). We need to calculate the shear stress on the pipe wall

Using formula of shear stress

[tex]\dfrac{\Delta p}{L}=\dfrac{2\tau}{r}[/tex]

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0.5}{2\times8\times12}[/tex]

[tex]\tau=0.2062\ lb/ft^2[/tex]

(b). We need to calculate the shear stress at the distance 0.3 in

Using formula of shear stress

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0.3}{2\times8\times12}[/tex]

[tex]\tau=0.12375\ lb/ft^2[/tex]

(c). We need to calculate the shear stress at the distance 0.5 in away from the pipe wall

r = 0.5-0.5 = 0

Using formula of shear stress

[tex]\tau=\dfrac{\Delta p\times r}{2L}[/tex]

Put the value into the formula

[tex]\tau=\dfrac{0.55\times144\times0}{2\times8\times12}[/tex]

[tex]\tau=0[/tex]

Hence, (a). The shear stress on the pipe wall is 0.2062 lb/ft²

(b). The shear stress at the distance 0.3 is 0.12375 lb/ft²

(c).  The shear stress at the distance 0.5 in away from the pipe wall is zero.

The electric potential at the midpoint between a proton and an electron is zero.

The electric potential at the midpoint between a proton and an electron can be calculated using the formula:

V = k * (q1 / r1 + q2 / r2)

where V is the electric potential, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r1 and r2 are the distances between the particles and the midpoint. In this case, the charges of the proton and the electron are equal in magnitude but opposite in sign, so q1 = -q2. The distances from the particles to the midpoint are equal because they are fixed in space, so r1 = r2. Plugging in the values, we get:

V = k * (-q / r1 + q / r1) = 0

Therefore, the electric potential at the midpoint between the two particles is zero.

Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

[tex]vf^{2} -vo^{2} = 2*a*x[/tex]

We know that v₀ = 0 (it starts from rest),  that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:

[tex]a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2[/tex]

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as  follows:

[tex]E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C[/tex]

⇒ E = 50.1*10³ N/C

Answer:

Explanation:

Maximum vertical distance or height = h = 35.4 m

let's consider the initial speed at the top is zero.

As the roller coaster is coming from top to bottom there is the conversion of gravitational potential energy into kinetic energy. So we will consider the law of conservation of energy.

As in this case,

Loss in potential energy = Gain in Kinetic energy

mgh = 1/2mv²

mass will cancel out will mass.

gh = 1/2 v²

v = √2gh

v = √2×9.8×35.4

v =√693.84

v = 26.34 m/s

The rollar coaster will have the maximum speed of 26.34 m/s when it reaches the bottom if we ignore the frictional forces.

Answer:

26.34 m/s.

Explanation:

Given:

h = 5.4 m

g = 9.81 m/s^2

Change in Potential energy = change in Kinetic energy

mgh = 1/2mv²

gh = 1/2 v²

v = √2gh

= √2×9.8×35.4

=√693.84

v = 26.34 m/s.

Angular displacement of the cd is 3.25 rev

Explanation:

The question is incomplete. It is not given what is the maximum angular velocity of the cd.

Here we are going to assume that the maximum angular velocity is:

[tex]\omega = 30 rad/s[/tex]

The motion of the cd is an accelerated angular motion, therefore we can use the following suvat equation:

[tex]\theta = (\frac{\omega_0 + \omega}{2})t[/tex]

where:

[tex]\theta[/tex] is the angular displacement of the cd during the time interval t

[tex]\omega_0[/tex] is the initial angular velocity of the cd

[tex]\omega[/tex] is the final angular velocity

Here we have:

t = 1.36 s

[tex]\omega_0 = 0[/tex] (assuming the cd starts from rest)

Therefore, the angular displacement of the cd during this time is:

[tex]\theta=(\frac{0+30}{2})(1.36)=20.4 rad[/tex]

And since [tex]1 rev = 2 \pi rad[/tex], we can convert into number of revolutions completed:

[tex]\theta = 20.4 rad \cdot \frac{1}{2\pi rad/rev}=3.25 rev[/tex]

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Explanation:

When the skydiver accelerates in the downward direction then tends to gain speed with each second. More is the resistance in air more will be the increase accompanied by the skydiver.

As a result, a point will come where air resistance force is balanced by gravitational force. Hence, the skydiver will attain terminal velocity.

So, air resistance for 1 [tex]m/s^{2}[/tex] = 0.45 [tex]m/s^{2}[/tex]

Air resistance for v [tex]m/s^{2}[/tex] = 0.45 v[tex]m/s^{2}[/tex]

As acceleration = change in velocity w.r.t time

                 a = [tex]\frac{dv}{dt}[/tex] = 0.45

            [tex]\frac{dv}{V}[/tex] = 0.45t

Now, we will integrate both the sides as follows.

            ln V = 0.45t

                V = [tex]e^{0.45t}[/tex]

Since,         [tex]F_{a} = F_{g}[/tex]   (in the given case)

so,                    ma = mg

On cancelling the common terms the equation will be as follows.

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

                         v = 21.77 [tex]m/s^{2}[/tex]

Thus, we can conclude that the skydivers terminal speed is 21.77 [tex]m/s^{2}[/tex].

Answer:

(a) 69.52 x 10⁻⁹ C

(b) 6.25 N/C

Explanation:

(a) The electric field (E) due to a point charge is directly proportional to the magnitude of the point charge (Q) and inversely proportional to the square of the distance (r) between the point charge and the point where the electric field is. This can be represented as follows;

E = k Q / r²         ---------------------------(i)

Where;

k = constant of proportionality called electric constant = 8.99 x 10⁹Nm²/C²

From the question, the following are given;

E = 10000N/C

r = 0.250m

Substitute these values into equation (i) as follows;

=> 10000 = 8.99 x 10⁹ x Q / 0.250²

=> 10000 = 8.99 x 10⁹ x Q / 0.0625

=> 10000 = 143.84 x 10⁹ x Q

Solve for Q;

Q = 10000 / (143.84 x 10⁹)

Q = 69.52 x 10⁻⁹ C

Therefore, the magnitude of the point charge is 69.52 x 10⁻⁹ C.

(b) By the same token, to calculate the magnitude of the electric field at 10.0m, substitute the values of Q = 69.52 x 10⁻⁹ C, k = 8.99 x 10⁹ and r = 10.0m into equation (i) as follows;

=> E = 8.99 x 10⁹ x 69.52 x 10⁻⁹ / 10.0²

=> E = 8.99 x 69.52 / 100.0

=> E = 6.25 N/C

Therefore, the electric field at 10.0m is as large as 6.25 N/C

Final answer:

The minimum separation r_min between two electrons moving towards each other and stopping due to repulsion is found by conserving mechanical energy, transforming the initial kinetic energy into electrostatic potential energy at the point of closest approach.

Explanation:

The minimum separation r_min that two electrons reach when moving directly toward each other and slowing down due to electric repulsion is determined by the conservation of energy. The total mechanical energy (kinetic plus potential) of the system must be conserved. At the maximum approach, both electrons will momentarily be at rest before repelling each other again, so all their kinetic energy is converted into electrical potential energy. The initial kinetic energy of both electrons, which is due to their motion, is transformed into electrostatic potential energy at the point of minimum separation.

The concept involves the calculation of this potential energy and setting it equal to the initial kinetic energy to find r_min. One might use the relationship between kinetic energy (1/2)m(v^2), electrostatic potential energy (k(q^2)/r), and the conservation of energy to solve for the unknown r_min. It's important to note that in such a scenario, relativistic effects are not considered as long as the velocity of the electrons is significantly less than the speed of light (c).

Answer:

Explanation:

Given

Spring constant of spring A is [tex]k_A=100\ N/m[/tex]

Spring constant of spring B is [tex]k_B=200\ N/m[/tex]

If displacement in both the springs is [tex]x=0.2\ m[/tex]

Potential Energy stored in the spring is given by

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

where k=spring constant

x=compression or extension

[tex]U_A=\frac{1}{2}\times 100\times (0.2)^2----1[/tex]

[tex]U_B=\frac{1}{2}\times 200\times (0.2)^2----2[/tex]

Divide 1 and 2

[tex]\frac{U_B}{U_A}=\frac{200}{100}=2[/tex]

[tex]U_A=\frac{U_B}{2}[/tex]

So Potential Energy Stored in Spring A is half of Spring B

The spring potential energy of spring A is half that of spring B, if the spring constant of spring A is 100 N/m and that of B is 200 N/m.

Spring constant of spring A, k₁ = 100 N/m

Spring constant of spring B, k₂= 200 N/m

Displacement of both the spring, x = 0.20 m

The potential energy stored in a spring is given by the equation U = 0.5kx²

For spring A, the potential energy is U₁ = 0.5 × 100×(0.20)² = 2 J.

For spring B, the potential energy is U₂ = 0.5 × 200×(0.20)² = 4 J

Therefore, the potential energy stored in spring A is half that of spring B.

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Answer:

The new force is 10 times the force F

Explanation:

Electric force between charged particles q1 and q2 at distance d is:

[tex]F=k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex] (1)

A new force between two different particles at a different distance is:

[tex]F'=k\frac{\mid q_{1}'q_{2}'\mid}{d'^{2}}=k\frac{\mid 5q_{1}\frac{q_{2}}{2}\mid}{(\frac{d}{2})^{2}}=\frac{5}{\frac{2}{4}}k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]

[tex]F'=10k\frac{\mid q_{1}q_{2}\mid}{d^{2}} [/tex]

Note that on the right side of the equation the term [tex]k\frac{\mid q_{1}q_{2}\mid}{d^{2}}=F [/tex] on (1), so:

[tex]F'=10F [/tex]

The force between the two charges, according to Coulomb's law, is approximately 1.92 Newtons. Note that the force is attractive since one charge is positive and the other is negative.

The question refers to Coulomb's law, which is fundamental in Physics. The law describes the force between two electric charges. We apply the formula |F|=K|QQ′|/d^2 where the charges are Q1=-13.5nC (converted to C by multiplying by 10^-9) and Q2=35.5nC, the distance d=1.735mm (converted to meters by multiplying by 10^-3), and K=1/4πϵ0≈8.99x10^9 N.m²/C². Inserting these values into the formula we get |F| ≈ 1.92 N. The magnitude of the force is therefore approximately 1.92 Newtons. It's important to note that since one charge is positive and the other negative, the force will be attractive, not repulsive.

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Answer:

The time taken for the monkey to reach the coconut, t is = H/v₀

Explanation:

Let the coordinate of the boy's hand be y = 0. The height of the tree above the boy's hand be H.

Coordinate of where the monkey meets coconut = y

Using the equations of motion,

For the monkey, initial velocity = 0m/s, time to reach coconut = t secs and the height at which coconut is reached = H-y

For the coconut, g = -10 m/s², initial velocity = v₀, time to reach monkey = t secs and height at which coconut meets monkey = y

For monkey, H - y = ut + 0.5gt², but u = 0,

H - y = 0.5gt²..... eqn 1

For coconut, y = v₀t - 0.5gt² ....... eqn 2

Substituting for y in eqn 1

H - y = H - (v₀t - 0.5gt²) = 0.5gt²

At the point where the monkey meets coconut, t=t

H - v₀t + 0.5gt² = 0.5gt²

v₀t = H

t = H/v₀

Solved!

a) Not possible

b) Yes, it's possible (see graph in attachment)

Explanation:

a)

The average velocity of a body is defined as the ratio between the displacement and the time elapsed:

[tex]v=\frac{\Delta x}{\Delta t}[/tex]

where

[tex]\Delta x[/tex] is the displacement

[tex]\Delta t[/tex] is the time elapsed

In this problem, we want to have zero displacement and non-zero average velocity. From the equation above, we see that this is not possible. In fact, if the total displacement is zero,

[tex]\Delta x = 0[/tex]

And therefore as a consequence,

[tex]v=0[/tex]

which means that the average velocity is zero.

B)

Here we want to have zero displacement and non-zero velocity. In this case, it is possible: in fact, we are not talking  about average velocity, but we are talking about (instantaneous) velocity.

On a position-time graph, the instantaneous velocity is the slope of the graph. Look at the graph in attachment. We see that the position of the object first increases towards positive value, then it decreases (the object starts moving backward), then becomes negative, then it increases again until returning to the original position, x = 0.

In all of this, we notice that the total displacement of the object is zero:

[tex]\Delta x = 0[/tex]

However, we notice that the instantaneous velocity of the object at the various instants is not zero, because the slope of the graph is not zero.

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Zero displacement and nonzero average velocity can occur on an x-t graph, but it is not possible to have zero displacement and nonzero velocity simultaneously.

Zero displacement and nonzero average velocity can occur when an object moves back and forth over a certain distance and returns to its original position, but with a change in time. For example, if you consider a person running in a circular track, after each lap the person returns to the starting point, resulting in zero displacement. However, if the person completes the laps in different time intervals, their average velocity will be nonzero.

On the other hand, it is not possible to have zero displacement and nonzero velocity simultaneously. Velocity is defined as the rate of change of displacement with respect to time. If the displacement is zero, then the velocity must also be zero

In an x-t graph, zero displacement would be represented by the horizontal line at the x-axis, or the line parallel to the x-axis. Nonzero average velocity would be represented by a sloped line that is above or below the x-axis, indicating movement in either the positive or negative direction.

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Answer:

(a) Surface charge density is the charge per unit area.

[tex]\sigma = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]

(b) [tex]\vec{E} = (+\^z)~8.34\times 10^3~{\rm N/C}[/tex]

(c) [tex]\vec{E} = (-\^z)~8.34\times 10^3~{\rm N/C}[/tex]

Explanation:

(a) Surface charge density is the charge per unit area. The area of the square plate can be calculated by its side length.

[tex]A = l^2 = (0.4)^2 = 0.16 ~{\rm m^2}[/tex]

Half of the total charge is distributed on one side and the other is distributed on the other side.

Therefore, surface charge density on each face of the plate is

[tex]\sigma = Q/A = \frac{2.35 \times 10^{-8}}{0.16} = 1.46 \times 10^{-7}~{\rm Q/m^2}[/tex]

(b) To find the electric field just above the plate, Gauss' Law can be used. Normally, Gauss' Law can only be used in infinite sheet (considering the flat surfaces), but just above the surface can be considered that the distance from the surface is much much smaller than the length of the plate (x << l).

In order to apply Gauss' Law, we have to draw an imaginary cylinder with radius r. The cylinder has to stay perpendicular to the plane.

[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r^2 = \frac{\pi r^2 \sigma}{\epsilon_0}\\E = \frac{\sigma}{2\epsilon_0}\\E = \frac{1.46 \times 10^{-7}}{2\times 8.8\times 10^{-12}} = 8.34\times 10^3~{\rm N/C}[/tex]

The direction of the electric field is in the upwards direction.

(c) The magnitude of the electric field is the same as that of upper side. Only the direction is reversed, downward direction.

Answer:

a) ΔV = 0.1875 V , b) v = 9.73 m / s

Explanation:

For this exercise we can use Necton's second law, as the speed is constant the forces on the driver are equal

               [tex]F_{E} - F_{B}[/tex] = 0

               [tex]F_{E}[/tex] = F_{B}

               q E = q v B

               E = v B

The electrical force induced in the conductor is

                ΔV = E l

                ΔV = v B l

Let's calculate

               ΔV = 2.5 0.75 0.10

               ΔV = 0.1875 V

b) If ΔV = 0.73 what speed should it have

             v = DV / B l

             v = 0.73 / 0.75 0.1

             v = 9.73 m / s

To find the volume of the snorkeler's lungs at a depth of 49m, use Boyle's Law by calculating the volume at the new pressure using the initial volume and pressure.

The volume of the snorkeler's lungs at a depth of 49 m can be calculated using Boyle's Law. Boyle's Law states that pressure and volume are inversely proportional when temperature is constant. To find the volume at depth, you can use P1V1 = P2V2, where P1V1 is the initial condition and P2V2 is the final condition.

Using the given data:

By rearranging the formula:

V2 = (P1 * V1) / P2 = (1.0 atm * 4.3 L) / 5.9 atm = 0.72 L

Therefore, at a depth of 49 m, the volume of the snorkeler's lungs would be approximately 0.72 liters.