At the instant when the electron is 4.40 cmcm from the wire and traveling with a speed of 6.10×104 m/sm/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron?

Answer

given,

displacement wave function

s(x, t) = 2.00 cos (15.7 x + 858 t)

now, comparing the wave equation with general equation.

s(x, t) = A cos (k x + ω t)

where A is the amplitude of the wave in micrometer.

now,

a) Amplitude of the wave

   A = 2 x 10⁻⁶ μ m

b) [tex]\lambda = \dfrac{2\pi}{k}[/tex]

       k = 15.7 m

   [tex]\lambda = \dfrac{2\pi}{15.7}[/tex]

             λ = 0.4 m

c) wave speed

    [tex]v = \dfrac{\omega}{k}[/tex]

    [tex]v = \dfrac{858}{15.7}[/tex]

           v = 54.65 m/s

d) For instantaneous displacement

Assuming the position and time is given as

x = 0.05 m and t = 3 m s

now,

s(x, t) = 2.00 cos (15.7 x 0.05 + 858 x 3 x 10⁻³ )

s(x, t) = 2.00 cos (3.359)

s(x,t) = -1.95 μ m

The amplitude of the wave is 2.00 micrometers, the wavelength is 15.7, and the speed of the wave is approximately 54.65 m/s.

The displacement wave function is given as s(x, t) = 2.00 cos (15.7x - 858t). In this wave function, the amplitude is the coefficient in front of the cosine function, which is 2.00 micrometers. The wavelength can be determined by finding the coefficient of the x term inside the cosine function, which is 15.7. The speed of the wave can be calculated by dividing the angular frequency (in this case, 858) by the wave number (in this case, 15.7), resulting in a speed of approximately 54.65 m/s.

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To solve this problem we will apply the definition of electrostatic force. From the variables present there and explained later we will find the value of the distance reorganizing said expression, that is

[tex]F = \frac{k q_1 q_2}{d^2}[/tex]

Here

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object

d = Distance

Replacing our values we have that

[tex]3 = \frac{(9*10^9)(1)(1)}{d^2}[/tex]

Rearranging and solving for the distance we have

[tex]d = 54772.25m[/tex]

Therefore the distance between the two objects is 54772.25m

Answer: %(∆W) = 0.37%

Explanation:

According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

F = Gm1m2/r^2

Where

F = force between the masses

G universal gravitational constant

m1 and m2 = mass of the two particles

r = distance between the centre of the two mass

Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth

W₁/W₂ = r₂²/r₁² .....1

W₂ = W₁r₁²/r₂²

At sea level the weight of the plane is W1 and at distance r₁ from the centre of the earth which is equal to the radius of the earth.

The radius of the earth is = 6378.1km

r₁ = radius of the earth = 6 378.1km = 6,378,100m

r₂ = r₁ + 11,719.342m = 6,378,100m + 11,719.342m

r₂ = 6,389,819.342m

W₂ = W₁r₁²/r₂²

W₂ = W₁[(6378100)²/(6,389,819.342)²]

W₂ = W₁[0.996335234422]

W₂/W₁ = 0.9963

fraction reduction of the weight is

ΔW/W₁ = 1 - W₂/W₁ = 1 - 0.9963 = 0.0037

percentage change :

%(∆W) = 0.0037 × 100% = 0.37%

Therefore, the percentage weight loss is 0.37%

Answer:

A) The energy stored in the capacitor increases.

Explanation:

For a capacitor fully charged by battery, when disconnected from battery and the plate separation is increased. The charge on the plate remain constant because there is no where for it to go( it has been disconnected from battery). But the capacitance would decrease, while also the potential difference would increase.

Q = CV ....1

Q is the charge, C is capacitance, V is the potential difference.

The energy stored in a capacitor is given by:

E = 1/2 CV^2

E = 1/2 QV .......2

E is the energy stored in the capacitor,

Therefore since Q remain constant and V increases when the distance between the plates is increased, then according to the equation 2 above the energy stored in the capacitor increases.

Answer:

34.35464 m/s

Explanation:

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

E = Electric field = 0.08 N/C

s = Displacement = [tex]4.2\times 10^{-8}\ m[/tex]

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

Electrical force is given by

[tex]F=qE[/tex]

Work done is given by

[tex]W=Fs\\\Rightarrow W=qEs[/tex]

Work done is also given by the kinetic energy

[tex]\dfrac{1}{2}mv^2=qEs\\\Rightarrow v=\sqrt{\dfrac{2qEs}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 0.08\times 4.2\times 10^{-8}}{9.11\times 10^{-31}}}\\\Rightarrow v=34.35464\ m/s[/tex]

The velocity of the electron is 34.35464 m/s

Answer:

23.0760769 %

Explanation:

[tex]m_1[/tex] = Mass of freight car = 15000 kg

[tex]m_2[/tex] = Mass of second car = 50000 kg

[tex]v_1[/tex] = Velocity of freight car = 2 m/s

[tex]v_2[/tex] = Velocity of second car = 0

v = Combined mass velocity

As the linear momentum of the system is conserved we have

[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]

The initial kinetic energy

[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]

Final kinetic energy

[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]

The percentage is given by

[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]

The change in percentage of initial kinetic energy is 23.0760769 %

Answer:

Can a room be gravitationally shielded? No, it can't.

Explanation:

the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.

Answer:

c.The stopping impulse is the same for either the hard objects or the airbag. Unlike the windshield or dashboard, the air bag gives some increasing the time for the slowing process and thus decreasing the average force on the passenger

Explanation:

The stopping impulse is the same for either the hard objects or the airbag.

Unlike the windshield or dashboard, the airbag gives some increasing the

time for the slowing process and thus decreasing the average force on the

passenger.

Air bags are put in automobiles ion order to prevent life threatening injuries

such as brain damage to individuals.  The air bag contains air helps

to increase the time for the slowing process.

This thereby prevents the average force on the passenger and prevents

injuries which should have occurred.

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Answer:

D and compound

Explanation:

because N2 is = to a compound

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

[tex]T_x = nsinA = \frac{mv^2}{r}[/tex]

[tex]T_y = ncosA = mg[/tex]

Dividing both.

[tex]tan A = \frac{v^2}{rg}[/tex]

[tex]tan A = \frac{11.7^2}{50*9.8}[/tex]

[tex]A = tan^{-1} (0.279367)[/tex]

[tex]A = 15.608\°[/tex]

Therefore the angle that should the curve be banked is 15.608°

Answer:

It takes 6.37 revolutions to stop.

Explanation:

The constant angular acceleration is negative if we choose the wheel direction of rotation as positive direction, so radial acceleration is α=5.0[tex] \frac{rad}{s^{2}}[/tex].Because the wheel is changing its velocity and we already know radial acceleration we should use the Galileo's kinematic rotational equation:

[tex]\omega^{2}=\omega_{i}^{2}+2\alpha\varDelta\theta [/tex]

with [tex] \omega[/tex] the final angular velocity (is zero because the wheel comes to rest), [tex] \omega_{i}[/tex] the initial angular velocity and Δθ the angular displacement. Solving (1) for Δθ :

[tex]\varDelta\theta=\frac{\omega^{2}-\omega_{i}^{2}}{2\alpha} [/tex]

[tex]\varDelta\theta=\frac{0-20^{2}}{(2)(-5.0)}=40rad [/tex]

The angular displacement can be converted to revolution knowing that 1 revolution is 2π rad:

[tex]40rad=\frac{40rad}{2\pi\frac{rad}{rev}} [/tex]

[tex] 40rad=6.37 rev[/tex]

Number of revolution take place is 6.4 revolution

Velocity of rotating wheel = 20 rad/s

Acceleration of magnitude = 5.0 rad/s²

Number of revolution take place

Using Third equation of motion;

20² - 0² = 2(5)(s)

400 = (10)(s)

Total distance = 40 rad

Number of revolution take place = Total distance / 2π

Number of revolution take place = 40 / 2(3.14)

Number of revolution take place = 40 / 6.28

Number of revolution take place = 6.369

Number of revolution take place = 6.4 revolution

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Answer:

Ex = -16.51 N/C Ey = 79.14 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1.

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward,(like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows.

E₁ = k*(3.65 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 8.99*10⁹*3.65*10⁻⁹ / (0.600)²m² = 91.15 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.10 m, 0.800 m) and (0,0), as follows:

r₂² = 1.10²m² + (0.800)²m² = 1.85 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 8.99*10⁹*(4.2)*10⁹ / 1.85 = 20.41 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.10 m / √1.85 m²) =- (1.10 / 1.36) = -0.809

By the same token, sin θ can be obtained as follows:

sin θ = - (0.800 m / 1.36 m) = -0.588

⇒E₂ₓ = 20.41 N/C * (-0.809) = -16.51 N/C (pointing to the left) (3)

⇒E₂y = 20.41 N/C * (-0.588) = -12.01 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-16.51 N/C) = -16.51 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  91.15 N/C + (-12.01 N/C) =79.14 N/C

Answer:

Ex = 15.505 N/C

Ey = 79.144 N/C

Explanation:

Particle 1

[tex]E_{1} = \frac{k*q_{1} }{r^2_{1}} \\\\r^2_{1} = 1.1^2+0.8^2\\\\r^2_{1} = 1.85 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.8}{1.1}) = 36.03 degree\\\\ E_{1} = \frac{(8.99*10^9)*(4.2*10^(-9)) }{1.85}\\\\E_{1} = 20.4096 N/C[/tex]

Away from the particle at (0,0) due to + charge

Particle 2

[tex]E_{2} = \frac{k*q_{2} }{r^2_{2}} \\\\r^2_{2} = 0.36 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.6}{0}) = 90 degree\\\\ E_{2} = \frac{(8.99*10^9)*(3.65*10^(-9)) }{0.36}\\\\E_{2} = 91.149 N/C[/tex]

Towards from the particle at (0,0) due to - charge

Resultant Electric field in y direction

[tex]E_{res,y} = E_{2} -E_{1}*cos(Q)\\E_{res,y} = (91.149) - (20.4096)*sin(36.03)\\\\E_{res,y} = 79.144 N/C[/tex]

Resultant Electric field in x direction

[tex]E_{res,x} = E_{1}*cos(Q)\\E_{res,y} =(20.4096)*cos(36.03)\\\\E_{res,x} = 16.505 N/C[/tex]

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of ​​the sphere

                I = P / A

               P = I A

The area of ​​the sphere is

               A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

               P = I₁ A₁ = I₂ A₂

               I₁ / I₂ = A₂ / A₁

 Suppose index 1 corresponds to the nearest planet,

            r2 = 2 r₁

            I₁ / I₂ = r₁² / r₂²

            I₁ / I₂ = r₁² / (2r₁)²

            I₁ / I₂ = ¼

           4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

[tex]m_{1}v_{1}=m_{2}v_{2}\\ v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s[/tex]

So Option d is correct one

The velocity of skater is [tex]0.07m/s[/tex]

Option d is correct.

For two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied.

                     [tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

                 [tex]65*v_{1}=0.15*32\\\\v_{1}=\frac{0.15*32}{65} =0.07m/s[/tex]

The velocity of skater is [tex]0.07m/s[/tex]

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The ratio between rotational energy and translational kinetic energy is [tex]1.47\cdot 10^{-3}[/tex]

Explanation:

The translational kinetic energy of the ball is given by:

[tex]KE_t = \frac{1}{2}mv^2[/tex]

where

m is the mass of the ball

v is the speed of the ball

The rotational kinetic energy of the ball is given by

[tex]KE_r = \frac{1}{2}I\omega^2[/tex]

where

I is the moment of inertia

[tex]\omega[/tex] is the angular speed

The moment of inertia of a solid sphere through its axis is given by

[tex]I=\frac{2}{5}mR^2[/tex]

where

m is the mass of the ball

R is its radius

Substituting into the previous equation,

[tex]KE_r = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2 \omega^2[/tex]

The ratio between the two energies is

[tex]\frac{KE_r}{KE_t}=\frac{\frac{1}{5}mR^2 \omega^2}{\frac{1}{2}mv^2}=\frac{2R^2 \omega^2}{5v^2}[/tex]

And substituting:

R = 3.91 cm = 0.0391 m

v = 33.6 m/s

[tex]\omega=52.1 rad/s[/tex]

we find:

[tex]ratio = \frac{2(0.0391)^2(52.1)^2}{5(33.6)^2}=1.47\cdot 10^{-3}[/tex]

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The ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere should be [tex]1.47.10^-3[/tex].

Since

The translational kinetic energy of the ball should be

[tex]KE_t = 1/2mv^2[/tex]

here

m is the mass of the ball

v is the speed of the ball

Now

The rotational kinetic energy of the ball should be

[tex]KE_r = 1/2Iw^2[/tex]

Here

I is the moment of inertia

w is the angular speed

Now

The moment of inertia of a solid sphere through its axis should be

[tex]I = 2/5mR^2[/tex]

Here

m is the mass of the ball

R is its radius

So,

[tex]KE_r = 1/2(2mR^2)w^2\\\\= 1/5mR62w^2[/tex]

Now

the ratio be

[tex]KE_r/KE_t = (1/2mR^2w^2) / (1/2mv^2)\\\\= (2R^2 w^2) / (5v^2)\\\\= 290.0391)^2 (52.1)^2 / 5(33.6)^2\\\\= 1.47*10^-3[/tex]

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The specific kinetic energy of a mass (in kilojoules per kilogram) moving at a velocity of 40 m/s is calculated using the kinetic energy formula, resulting in 0.8 kJ/kg.

The specific kinetic energy of a mass moving with a velocity can be determined using the formula for kinetic energy, K = 0.5*m*v². In this case, where the velocity 'v' is given as 40 m/s, and we want to solve the kinetic energy per kilogram, we can consider the mass 'm' as 1 kg. Hence the specific kinetic energy would be K = 0.5*(1 kg)*(40 m/s)² = 800 J = 0.8 kJ/kg, because 1 kilojoule (kJ) = 1000 joules (J).

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Answer:

A) At 2.2 s the position of the particle is 9.4 m.

B) At t =2.2 s the velocity is 9.7 m/s.

C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²

Explanation:

Hi there!

A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:

dx/dt = 2 · t²

Separating variables, we can find the equation of position as a function of time:

dx = 2 · t² · dt

Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:

∫ dx = 2 ∫ t² · dt

x - 2.3 m = 2/3 · t³

x = 2.3 m + 2/3 m/s³ · t³

Replacing t = 2.2 s:

x = 2.3 m + 2/3 m/s³ · (2.2 s)³

x = 9.4 m

At 2.2 s the position of the particle is 9.4 m

B) Now, let´s evaluate the velocity function at t = 2.2 s:

v = 2 · t²

v = 2 m/s³ · (2.2 s)²

v = 9.7 m/s

At t =2.2 s the velocity is 9.68 m/s

C) The acceleration is the variation of the velocity over time (the derivative of the velocity):

dv/dt = a

a = 4 · t

At t = 2.2 s:

a = 4 m/s³ · 2.2 s

a = 8.8 m/s²

At t = 2.2 s the acceleration of the particle is 8.8 m/s²

(A) The particle's position at time, t = 2.2 s is 7.1 m.

(B) The velocity of the particle at 2.2 s is 9.68 m/s.

(C) The acceleration of the particle at 2.2 s is 8.8 m/s².

The given parameters:

The particle's position at time, t = 2.2 s is calculated as follows;

[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]

The velocity of the particle at 2.2 s is calculated as follows;

[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]

The acceleration of the particle at 2.2 s is calculated as follows;

[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]

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Explanation:

Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.

Explanation:

Water in the air (humidity) helps to dissipate static charge that builds up. If the air is very dry, the charge can't dissipate, so it builds up until there is enough to spark.

Tropical countries are typically more humid than the United States, but I guess that depends on where you are in the US.

A student's first experience with static electricity in an American winter is due to the dry air, which allows for greater accumulation and discharge of electrical charges, unlike in humid tropical climates.

This is because static electricity is more prevalent in cold, dry environments.

In summary, the primary reason a student may experience static electricity for the first time in an American winter is due to the dry air conditions that favour the accumulation and sudden discharge of electric charges.

The maximum Volume Flow Rate of water the pump can provide, given an efficiency of 82% and an elevation of 15 m, is approximately 0.033 L/s.

First, we must convert the pump's horsepower to a more usable unit in this context - like watts. In physics, 1 horsepower equals roughly 746 watts. Therefore, the pump has power of 8*746 = 5968 watts.

Given the mechanical efficiency (ME) and the height (h), the maximum work the pump can do is given by M.E. * Power. So, the pump does work of 0.82*5968 = 4895.76 watts.

The work done on the water by the pump is equal to the change in potential energy of the water, PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15 m). With rearranging, you could express m = Power/(g * h). But we're looking for the volume flow rate, not the mass flow rate, so we need to convert mass to volume. Since the density of water (ρ) is 1 kg/L, the volume flow rate = m/ρ = Power/(g * h * ρ).

Substituting all known values, we get: Volume flow rate = 4895.76 W /(9.8 m/s^2 * 15 m * 1 kg/L) = 0.033 L/s.

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To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,

[tex]x = \frac{\sqrt{3}}{2}a[/tex]

[tex]x = \frac{\sqrt{3}}{2}(0.5)[/tex]

[tex]x = 0.43301m[/tex]

The magnitude of the electric field is

E=\frac{kQ}{x^2}

Here,

k = Coulomb's Constant

Q = Charge

x = Distance

[tex]E=\frac{(9*10^9)(5*10^{-9})}{(0.43301)^2}[/tex]

[tex]E = 240.002N/C \approx 240N/C[/tex]

Therefore the correct answer is A.

The new temperature is: d. 200 K

The Ideal Gas Law is given by:

[tex]PV = nRT[/tex]

where:
[tex]P[/tex] = pressure
[tex]V[/tex] = volume
[tex]n[/tex] = number of moles
[tex]R[/tex] = universal gas constant
[tex]T[/tex] = temperature

Given initial conditions:

[tex]P_1 = 2[/tex] atm (gauge pressure)
[tex]T_1 = 200[/tex] K
[tex]V_1 = V[/tex]
[tex]n = 10[/tex] moles

Final conditions:

[tex]P_2 = 1[/tex] atm (gauge pressure)
[tex]V_2 = 2V[/tex]
[tex]T_2 = ?[/tex]

We can use the combined gas law equation to relate the initial and final states of the gas:

[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]

Substituting in the given values:

[tex]\frac{2V}{200} = \frac{1 \cdot 2V}{T_2}[/tex]

Solving for [tex]T_2[/tex]:

[tex]\frac{2V}{200} = \frac{2V}{T_2}[/tex]

By simplifying, we get:

[tex]\frac{2}{200} = \frac{2}{T_2}[/tex]

Cross-multiplying gives:

[tex]2 \cdot T_2 = 400[/tex]

Dividing both sides by 2 gives:

[tex]T_2 = 200[/tex] K

Therefore, the new temperature [tex](T_2)[/tex] after the volume is doubled and the pressure is halved is 200 K.

To solve this problem we will use the concepts given by Coulomb's law defined for force, said law is mathematically described as

[tex]F = \frac{kq_1 q_2}{r^2}[/tex]

Here,

k = Coulomb's constant

[tex]q_{1,2}[/tex]= Charge of two protons

r = Distance between them

F = Force

Our values are given as,

[tex]F =2.3*10^{-26} N[/tex]

[tex]q_1 =q_2 = 1.6*10^{-19} C[/tex]

[tex]k =9*10^9 Nm^2/C2[/tex]

Rearrenging to find the distance and replacing we have that

[tex]r^2=\frac{(9*10^9 )(1.6*10^{-19})^2 }{2.3*10^{-26} }[/tex]

[tex]r^2=10.01*10^{-3} m^2[/tex]

[tex]r =\sqrt{10.01*10^{-3} m^2}[/tex]

[tex]r = 0.100m[/tex]

Therefore the correct option is B.

Answer:

(a). The final velocity is 1.78 m/s.

(b). The lost kinetic energy is 266.13 J.

Explanation:

Given that,

Mass of first skater = 60 kg

Mass of second skater = 75.0 kg

Initial velocity = 4.00 m/s

(a). We need to calculate the final velocity

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v[/tex]

Put the value into the formula

[tex]60\times4.00+75.0\times0=(60+75.0)v[/tex]

[tex]v=\dfrac{60\times4.00}{60+75}[/tex]

[tex]v=1.78\ m/s[/tex]

The final velocity is 1.78 m/s.

(b). We need to calculate the lost kinetic energy

Using formula of kinetic energy

[tex]\Delta E=E_{2}-E_{1}[/tex]

[tex]\Delta E=\dfrac{1}{2}(m_{1}u_{1}^2+m_{2}u_{2}^2)-\dfrac{1}{2}(m_{1}+m_{2})v^{2}[/tex]

[tex]\Delta E=\dfrac{1}{2}(60\times4^2)-\dfrac{1}{2}\times(60+75)\times1.78^2[/tex]

[tex]\Delta E=266.13\ J[/tex]

Hence, (a). The final velocity is 1.78 m/s.

(b). The lost kinetic energy is 266.13 J.

a. The final velocity should be considered as the 1.78 m/s.

b. The lost kinetic energy  should be considered as the 266.13 J.

Since

Mass of first skater = 60 kg

Mass of second skater = 75.0 kg

Initial velocity = 4.00 m/s

a.

We know that

[tex]mm1v1 + m2v2 = (m1 + v1)v\\\\60*4.00 + 75*0 = (60 + 75)v[/tex]

v = 60 + 4/60 + 75

= 1.78 m/s

b.

[tex]E = 1/2(m1v1^2 + m2v2^2) - 1/2(m1 + vm2)v^2\\\\= 1/2(60*4^2) - 1/2 * (60 + 75)*1.78^2[/tex]

= 266.13 J

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Answer:

[tex]F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

[tex]F_H_n=F_f\\F_h=mg*u[/tex]

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

[tex]200=mg*u[/tex]                         Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

[tex]F_H=(1.42m*g*.81u)[/tex]          Eq (2)

Divide Eq(2) and Eq (1)

[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]

Explanation:

A vector in this situation have two components

1) The X-component

2) The Y-component

So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be

[tex]A_x[/tex] = Acos[tex]\theta[/tex]

I hope this will answer your question,

An image is also provided please have a look at that.

Thank you.

Answer: Ax = Acos(theta)

Therefore, the x component of A: Ax = Acos(theta)

Explanation:

The attached image is a pictorial representation of the question:

From the attached image,

Cos(theta) = adjacent/hypothenus

Cos(theta) = Ax/A

Making Ax the subject of formula,

Ax = Acos(theta)

Therefore, the x component of A: Ax = Acos(theta)

Answer:

[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R

Explanation:

If we define the magnitude of the field as B, then we have:

Total magnitude of the field [tex]B_{t}[/tex] = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.

B_loop = (μ0)*([tex]I_{2}[/tex])/2*R

B_wire = (μ0)*([tex]I_{1}[/tex])/2*π*D

Thus:

B_loop = B_wire at the center of the loop.

(μ0)*([tex]I_{2}[/tex])/2*R = (μ0)*([tex]I_{1}[/tex])/2*π*D

[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R

The magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].

The given parameters;

The magnetic field at the center of the loop is calculated as follows;

[tex]B_o = \frac{\mu_o I_2}{2R}[/tex]

The magnetic field at the distance of the wire is calculated as follows;

[tex]B_o = \frac{\mu_o I_1}{2\pi D}[/tex]

The magnitude of current at the center of the loop is calculated as follows;

[tex]\frac{\mu_o I_2 }{2R} = \frac{\mu_o I_1}{2\pi D} \\\\I_1 = \frac{\pi D I_2}{R}[/tex]

Thus, the magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].

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a) The minimum frequency of the light must be [tex]1.26\cdot 10^{15} Hz[/tex]

b) The maximum velocity of the electrons is [tex]5.93\cdot 10^5 m/s[/tex]

Explanation:

a)

The photoelectric effect is a phenomenon that occurs when electromagnetic radiation hits the surface of a metal causing the release of electrons from the metal's surface.

The equation of the photoelectric effect is:

[tex]hf = \phi +K_{max}[/tex]

where :

[tex]hf[/tex] is the energy of the incoming photons, where

[tex]h[/tex] is the Planck's constant

[tex]f[/tex] is the frequency of the incoming photons

[tex]\phi[/tex] is the work function of the metal, the minimum energy that the photons must have in order to be able to free electrons from the metal

[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted electrons

In order to free electrons, the minimum energy of the photons must be at least  equal to the work function (so that the kinetic energy of the electrons is zero, [tex]K_{max}=0[/tex]. Therefore,

[tex]h f_0 = \phi[/tex]

In this case,

[tex]\phi = 5.22 eV \cdot (1.6\cdot 10^{-19})=8.35\cdot 10^{-19} J[/tex]

Therefore, the minimum frequency of the photons must be

[tex]f_0 = \frac{\phi}{h}=\frac{8.35\cdot 10^{-19}}{6.63\cdot 10^{-34}}=1.26\cdot 10^{15} Hz[/tex]

b)

In this case, the wavelength of the incoming light is

[tex]\lambda = 200 nm = 200 \cdot 10^{-9} m[/tex]

We can find the frequency by using the wave equation:

[tex]f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{200\cdot 10^{-9}}=1.5\cdot 10^{15} Hz[/tex]

Now we can use the equation of the photoelectric effect to find the maximum kinetic energy of the electrons:

[tex]K_{max} = hf-\phi = (6.63\cdot 10^{-34})(1.5\cdot 10^{15})-8.35\cdot 10^{-19}=1.60\cdot 10^{-19} J[/tex]

And therefore, we can find their velocity by using the equation for the kinetic energy:

[tex]K_{max} = \frac{1}{2}mv^2[/tex]

where

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electrons

v is their speed

Solving for v,

[tex]v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})}{9.11\cdot 10^{-31}}}=5.93\cdot 10^5 m/s[/tex]

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Final answer:

Each consecutive squirt from the water gun is separated by 3 meters, calculated by multiplying the speed of the squirts (15 m/s) by the interval between each squirt (0.2 seconds).

Explanation:

The question is asking to calculate the distance at which each consecutive squirt from a water gun is separated when the water gun fires squirts at a certain rate and speed. Given that the water gun fires 5 squirts per second and the speed of the squirts is 15 m/s, we can use the formula distance = speed × time to find the separation distance between squirts.

Since there are 5 squirts per second, each squirt is separated by 1/5 of a second, or 0.2 seconds. To find the separation distance, we multiply the speed of the squirts by the time interval between each squirt:

Distance = Speed × Time
Distance = 15 m/s × 0.2 s
Distance = 3 meters

Therefore, each consecutive squirt is separated by 3 meters.