The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 1 mph , toward a group of small children. Seeing the situation, you realize you have just enough time to drive your 2000 kg Volkswagen head-on into the Cadillac and save the children. You may want to review (Page 269) . Part A With what speed should you impact the Cadillac to bring it to a halt?

Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

Answer:

The minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm.

Explanation:

The diameter of person's eye is 2.5 cm. The close point or the near point of the eye is 25 cm and the far point is infinity. We need to determine the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye. Let f is the minimum effective focal length. It can be calculated using lens formula as :

[tex]\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{2.5}[/tex]

f = 2.27 cm

So, the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm and it is at the nearest point. Hence, this is the required solution.

Answer:

The speed of the particle after travelling for 0.5 m is 100 m/s.                            

Explanation:

It is given that,

Mass of the particle, m = 0.01 kg

Net charge on the particle, q = -0.05 C  

Electric field strength, E = 2000 V/C

Distance travelled by the particle, d = 0.5 m

The work done due to motion of the particle is balanced by the change in kinetic energy as :

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

v is the speed of the particle

F is the electric force

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

[tex]v=\sqrt{\dfrac{2qEd}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}[/tex]

v = 100 m/s

So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.

Answer:

24 hours

[tex]7\times 10^{-5}\ rad/s[/tex]

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

[tex]T=24\times 60\times 60\ s[/tex]

Angular velocity is given by

[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s[/tex]

The angular velocity of the satellite is [tex]7\times 10^{-5}\ rad/s[/tex]

A communications satellite in a geostationary orbit must have a period of 24 hours and an angular velocity of approximately 0.262 radians per hour.

To remain above the same point on Earth's surface, a communications satellite must be placed in a geostationary orbit. The period of such a satellite must be 24 hours, which is the same as one Earth day. This means that the satellite completes one orbit around the Earth in 24 hours. The angular velocity of the satellite depends on its position in the orbit and can be calculated using the formula:

Angular velocity = 2π / Period

Since the period is 24 hours, the angular velocity of the satellite is approximately 0.262 radians per hour.

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The spring constant of the spring is approximately 234.50 N/m.

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:

F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m

Therefore, the spring constant of the spring is approximately 234.50 N/m.

Answer:

 a₂ = m₁ / m₂ a₁

Explanation:

For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies

Let's apply Newton's second law on the star 1

     F₁ = m₁ a₁

Newton's second law in star 2

       F₂ = m₂ a₂

       | F₁ | = | F₂ |

      m₁  a₁ = m₂  a₂

      a₂ = m₁ / m₂ a₁

Answer:

a. get warmer.

Explanation:

When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.

It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.

Answer:

[tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

Explanation:

Light which travels from the crackers reaches the detector at [tex]c=3\times 10^{8}\ m/s[/tex]

[tex]\Delta x_1[/tex] = Distance at which event 1 leaves a char mark = 3.4 m

[tex]\Delta x_2[/tex] = Distance at which event 2 leaves a char mark = 2.1 m

The speed of light in a medium is a universal constant

[tex]c=\dfrac{\Delta x_1}{\Delta t_1}\\\Rightarrow \Delta t_1=\dfrac{\Delta x_1}{c}\\\Rightarrow \Delta t_1=\dfrac{3.4}{3\times 10^8}\\\Rightarrow \Delta t_1=1.13333\times 10^{-8}\ s[/tex]

[tex]c=\dfrac{\Delta x_2}{\Delta t_2}\\\Rightarrow \Delta t_2=\dfrac{\Delta x_2}{c}\\\Rightarrow \Delta t_2=\dfrac{2.1}{3\times 10^8}\\\Rightarrow \Delta t_2=0.7\times 10^{-8}\ s[/tex]

The pulse will be detected at [tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]

In the reference frame of the detector, both light pulses will be detected approximately 1.13 x 10^-8 seconds after they occur.

In the reference frame of the detector, event 1 leaves a mark at a distance of 3.40 m and event 2 leaves a mark at a distance of 2.10 m. Since the events are simultaneous in the reference frame of the detector, the time it takes for each light pulse to be detected will be the same for both events. To calculate the time, we can use the speed of light as the distance traveled divided by the speed of light. Therefore, the time it takes for each light pulse to be detected is t = distance/speed of light.

For event 1, the distance is 3.40 m and for event 2, the distance is 2.10 m. The speed of light is approximately 3.00 x 10^8 m/s. Plugging in these values, we get:

t = (3.40 m) / (3.00 x 10^8 m/s) = 1.13 x 10^-8 s

Therefore, each light pulse will be detected approximately 1.13 x 10^-8 seconds after the events occur.

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Answer:

0.1587

Explanation:

Given data

μ = 0.5 mg  

standard deviation σ =0.1 mg

n=100

we know that

[tex]P(\overline X<0.49)[/tex]

[tex]Z= (\frac{\overline X-\mu}{\sigma/\sqrt{n} })[/tex]

putting the values we get

[tex]Z= (\frac{0.49-0.5}{0.1/\sqrt{100} })[/tex]

Z=-1

Area under the curve for z =-1 is 0.1587 (from z score table)

P(X<0.49) = 0.1587

P(X<0.37) = 0.0013

Answer:

Explanation:

Given

density of cylinder is [tex]\rho _c=713 kg/m^3[/tex]

Length of first cylinder is [tex]L_1=20 cm[/tex]

radius [tex]r_1=5 cm[/tex]

For cylinder 2 [tex]L_2=10 cm [/tex]

[tex]r_2=10 cm[/tex]

[tex]h_1[/tex] and [tex]h_2[/tex] are the height above water

E

as object is floating so its weight must be balanced with buoyant force

[tex]\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1[/tex]

For 2nd cylinder

[tex]\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2[/tex]

Dividing 1 and 2 we get

[tex]\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}[/tex]

[tex]\frac{20}{10}=\frac{20-h_1}{10-h_2}[/tex]

[tex]2h_2=h_1[/tex]

[tex]\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}[/tex]                            

Answer:

[tex]1.04\times 10^7\ J.[/tex]

Explanation:

In the question given :

Pressure is constant

Therefore, Work done, [tex]W=P\times\Delta V[/tex]

Pressure, P=1.01 × 105 Pa.

Final volume, [tex]V_f=8.50\ m^3.[/tex]

Initial volume, [tex]V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.[/tex]

Therefore, W=8.58\times 10^{5}\ J.

Also, Heat Given, [tex]Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.[/tex]

Also, according to First law of thermodynamics:

[tex]\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.[/tex]

Hence, this is the required solution.

The energy needed to convert all of the water into steam in this isothermal system is 11.30 MJ. This is calculated using the formula Q = mLv, substituting the given values for mass and latent heat of vaporization.

The heat required to convert water from liquid into steam, in an isothermal process, can be found using the formula Q = mLv, where 'm' is the mass of the water, and 'Lv' is the latent heat of vaporization.

Given that we have 5.00 kg of water and the heat of vaporization for water under atmospheric pressure is 2.26 × 106 J/kg, we can substitute these values into our formula:

Q = 5.00 kg × 2.26 × 106 J/kg.

This gives us Q = 11.30 × 106 J, or 11.30 MJ. Therefore, 11.30 MJ of heat is added to the system to convert all the water into steam at atmospheric pressure.

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The sets of driver's actions “Signalling; Changing Gears; Checking mirrors” are multi-task performances necessary for the safe vehicle operation.

Answer: Option A

Explanation:

Signaling is what drivers plan to do is important for security because other drivers can only know if he tells them. Here are some general rules about Signalling or turns.

The correct gear change is important. If he can't put the vehicle in the right gear while driving, he has less control. And also, he must check the exterior mirrors on both sides regularly. He needs to check his vehicle mirrors to make sure nobody stands or walks past him.

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

The rate at which the depth of the water is changing is -0.178 cm/sec, and the rate at which the radius of the top of the water is changing is -0.14 cm/sec.

The subject of this question is related to Calculus and the specific topic is related rates. For this problem, in addition to the rate at which water is leaking, we need to consider the geometric relationship within the cone-shaped water tank.

First, we are given that V' = -35 cm^3/sec (we make it negative because the volume is decreasing) and we know that the volume of a cone is V = (1/3)πr²h. Our tank parameters are r = 1m and h = 2.5m, but we need everything in the same units, so we convert our radius to 100 cm.

We know through similar triangles that r/h = R/H, where r and R are the radii at any given time, and h and H are the heights at any given time respectively. Thus, r = (Rh)/H. Substituting r in our volume equation, we get: V = (1/3)π((Rh)^2)/H=h²πR²/H, and hence V=hπR². Differentiating this implicitly with t gives V' = h'πR²+2hπRr'.

Substituting for V', h and r from our given information, we get: -35=h'πR²+2(1.25)πRr'. We can solve for h' and r' separately.

(a) To find h' we set r' = 0, because we only want to know how depth h is changing. Solving for h' we find it to be -0.178 cm/sec.

(b) To find r', we set h' = 0, because we are only interested in how radius r is changing. Solving for r' we get -0.14 cm/sec.

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Answer:

c

Explanation:

from. 50 to 60 diopters

The range of Anna's vision will be between 0-5 diopters.

It is given that

Anna is able to focus on objects within 20 cm of her eyes (her near point)

Now the distance between the cornea and retina, of Anna's eye is 20 mm

The focal length will be f= 20+2= 22cm

Now the power of eyes will be given by

[tex]P= \dfrac{1000}{f(in \ mm)}[/tex]

[tex]P= \dfrac{1000}{ 22}=4.54 \ diopters[/tex]

Thus the range of Anna's vision will be between 0-5 diopters.

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Answer:

-0.23694 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

dr = Height = 1 mile = 1609.34 m

Acceleration is given by

[tex]a=\dfrac{GM}{r^2}[/tex]

Change in acceleration is given by

[tex]da=-2\dfrac{GM}{r^3}dr[/tex]

[tex]w=ma\\\Rightarrow w=m\dfrac{GM}{r^2}\\\Rightarrow w=469\ N[/tex]

[tex]dw=mda\\\Rightarrow dw=-m2\dfrac{GM}{r^3}dr\\\Rightarrow dw=-2w\dfrac{dr}{r}\\\Rightarrow dw=-2\times 469\times \dfrac{1609.34}{6.371\times 10^{6}}\\\Rightarrow dw=-0.23694\ N[/tex]

The change in weight is -0.23694 N

The change in your weight if you were to ride an elevator from the street level where you weigh 469N to the top of the building is; -0.237 N

The formula for acceleration here is;

a = GM/r²

Where;

G is gravitational constant = 6.67 × 10⁻¹¹ m³/kg.s²

M is mass of earth = 5.972 × 10²⁴ kg

r is distance from center of earth

Since we are trying to find change in weight, let us first find the change in acceleration with respect to r;

da/dr = -2GM/r³

da =  -(2GM/r³) dr

Thus, change in weight from top to bottom is;

W_top - W_bottom = m(da)

Now, weight at bottom is gotten from the formula;

W_bottom = GmM/r²

Also, W_bottom = m(da) since we are dealing with change in weight.

Thus;

m(da)= -(2GmM/r³) dr

Recall that GmM/r². Thus;

m(da) = -2W_bottom × dr/r

where;

W_bottom = 469 N

r is radius of earth = 6371000 m

dr = 1 mile = 1609.34 m

Thus;

m(da) = -2 × 469 × 1609.34/6371000

m(da) = -0.237 N

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Answer:

a) -6,750 N  b) -1.35*10⁶ N  c) 11.5 times d) 2,295 times

Explanation:

a) In order to answer the question, we can apply Newton's 2nd Law to the person, as follows:

Fnet = m*a

Assuming that the deceleration is uniform, we can use any of the kinematic equations.

In this particular case, examining the givens that we have (the final speed (which is 0), the initial speed (impact speed), and the distance of deceleration (1 m) the most useful equation is the following:

vf²-v₀² = 2*a*d

Replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (1 m) = -112.5 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-112.5 m/s²) = -6,750 N

b) For this part, same reasoning applies, the only difference being the deceleration distance, which for this case is only 5 mm.

We can apply the same kinematic equation:

vf²-v₀² = 2*a*d

Once again, replacing by the givens, and solving for a, we get:

a = (15m/s)² / 2* (0.005 m) = -22,500 m/s²

With this value of a, we can get the net force F:

F = 60 kg* (-22,500 m/s²) = -1.35*10⁶ N

c) If we compare the forces that  we got above with the weight of the person (which is the same to compare the acceleration with g), we have, for the first case (person restrained) a value of  approximately 12 g, but for the unrestrained case, we got a value of 2,295 g!

Answer:

U. With no variation.

Explanation:

Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

Answer:

1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.

2- Similarly, T3 = T.

3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is

[tex]v = \omega R[/tex]

For all the people, angular velocity is the same. Their linear velocities are different.

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]

4- Similarly,

[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]

5- Radial acceleration in constant circular motion is

[tex]a_{rad} = \frac{v^2}{R}[/tex]

For the second person:

[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]

6- Similarly,

[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]

Explanation:

As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.

The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.

In the case where the angular accleration is zero, the expression for the velocity is:

        [tex]w = \frac{\Delta \theta }{\Delta t}[/tex]  

Where w is the angular velocity and Δw and Δt are the variation in angle t over time.

1 and 2)

Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.

          [tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]  

In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.

         T = T₁ = T₂

3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position  r = [tex]\frac{3}{5} \ R[/tex]  

Linear and angular variables are related.

        v = w r

Let's substitute for the 1st person.

        v = w R

For the 2nd person.

        v₂= w ( [tex]\frac{3}{5} R[/tex])

We solve these two equations.

         [tex]v_2 = \frac{3}{5} \ v[/tex]  

4) We carry out the same calculation for the 3rd person.

         v₃ = w ½ R

We solve the two equations.

          v₃ = ½ v

5) Ask for radial acceleration.  

The relationship between radial and angular acceleration is.

           a = α R

We substitute for the 1st person.

          a = α R

For the second person.

          a₂ = α ( [tex]\frac{3}{5} R[/tex])

We solve the two equations

         a₂ = [tex]\frac{3}{5} \ a[/tex]

6) Ask the radial acceleration of the 3rd person.

We substitute.

         a₃ = α (½ R)

We solve.

        a₃ = ½ a

In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are

      1 and 2) All periods are equal, T₂ = T and T₃ = T.

       3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]  

       4) The linear velocity of the 3rd person is: v₃ = ½ v

       5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]  

      6) The linear acceleration of the 3rd person: a₃ = ½ a

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Answer:

1350N

Explanation:

[tex]2.75 ms = 2.75*10^{-3}s[/tex]

The force exerted on the hand would be the momentum divided by the duration of contact.

As the hand is coming to rest, final velocity would be 0

[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]

The magnitude of the force would be 1350N

Final answer:

The magnitude of the electric field just outside the surface of a sphere designed to deflect solar radiation on the moon or Mars is 10⁷ N/C, calculated based on the given requirements for the electric field at a distance from the sphere.

Explanation:

The question asks for the magnitude of an electric field E just outside the surface of a sphere that is proposed to protect astronauts on the surface of the moon or Mars by deflecting high-energy charged particles emitted by the sun.

The sphere, assumed to have a uniform surface charge distribution, generates an electric field whose magnitude at a distance r from its center can be found using the formula E = kQ/r², where k is Coulomb's constant (8.99 x 10⁹ N·m²/C²), Q is the charge on the sphere, and r is the distance from the center of the sphere.

Given that the sphere's electric field's magnitude at a distance of 25 m from its center needs to be 1 x 10⁶ N/C, we find the charge Q required to produce this field.

Once Q is determined, we can calculate the electric field's magnitude just outside the sphere's surface (at r = 2.5 m, which is the radius of the 5 m diameter sphere) using the same formula.

The calculation reveals that the magnitude of E just outside the surface of the sphere is of the order 10⁷ N/C, making option c) 10⁷ N/C the correct answer.

The magnitude of the electric field just outside the surface of the sphere is [tex]\(10^7 \, \text{N/C}\),[/tex] so the correct answer is (c) [tex]\(10^7 \, \text{N/C}\).[/tex]

To find the magnitude of the electric field just outside the surface of the sphere, we'll use Gauss's Law, which states:

[tex]\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\][/tex]

Where:

- \(\vec{E}\) is the electric field vector

- \(d\vec{A}\) is a differential area vector

- \(Q_{\text{enc}}\) is the enclosed charge

- \(\varepsilon_0\) is the permittivity of free space

Since the electric field is radial, we can express it as [tex]\(E = E(r)\hat{r}\)[/tex], where [tex]\(\hat{r}\)[/tex] is a unit vector pointing radially outward from the center of the sphere.

The magnitude of the electric field \(E\) just outside the surface of the sphere is equal to the magnitude of the electric field produced by a point charge at the center of the sphere, which is given by Coulomb's Law:

[tex]\[E = \frac{k |Q|}{r^2}\][/tex]

Where:

- \(k\) is Coulomb's constant [tex](\(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\))[/tex]

- \(|Q|\) is the magnitude of the charge enclosed by the Gaussian surface

- \(r\) is the distance from the center of the sphere

Given that [tex]\(E = 1 \times 10^6 \, \text{N/C}\) at \(r = 25 \, \text{m}\), we can solve for \(|Q|\):[/tex]

[tex]\[1 \times 10^6 \, \text{N/C} = \frac{k |Q|}{(25 \, \text{m})^2}\][/tex]

[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (25 \, \text{m})^2}{k}\][/tex]

[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (625 \, \text{m}^2)}{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2}\][/tex]

[tex]\[|Q| = \frac{625 \times 10^6}{8.99}\][/tex]

[tex]\[|Q| \approx 69.633 \, \text{C}\][/tex]

Now, let's calculate the electric field just outside the surface of the sphere using this charge:

[tex]\[E = \frac{k |Q|}{r^2}\][/tex]

[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{(5 \, \text{m})^2}\][/tex]

[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{25 \, \text{m}^2}\][/tex]

[tex]\[E = \frac{629.586 \times 10^9}{25}\][/tex]

[tex]\[E \approx 25.183 \times 10^7 \, \text{N/C}\][/tex]

Therefore, the magnitude of the electric field just outside the surface of the sphere is approximately [tex]\(2.5183 \times 10^8 \, \text{N/C}\).[/tex]

The closest option is d) [tex]\(10^8 \, \text{N/C}\).[/tex]

Answer:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop

Explanation:

The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines.

Mathematically:

[tex]\phi_B=\vec{B}.\vec{A}[/tex]

where:

[tex]\vec{A}=[/tex] area vector directed normal to the surface

[tex]\vec{B}=[/tex] magnetic field vector

Final answer:

Rotating the conducting loop about an axis in the z direction will not change the magnetic flux through the loop because the angle between the magnetic field and the normal to the plane of the loop remains constant. so the correct option is D

Explanation:

The question pertains to the change in magnetic flux through a conducting loop when subjected to different actions. According to Faraday's Law of Electromagnetic Induction, the magnetic flux through a loop is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field direction and the normal to the loop. In options A, B, and C, changes to the area or the strength of the magnetic field alter the magnetic flux since these factors directly affect the calculation of flux. Option D involves rotating the loop about an axis in the z direction; such a rotation does not change the angle between the magnetic field and the normal to the loop's plane, therefore would not change the flux. In contrast, option E, where the loop is rotated about an axis in the y direction, changes this angle and thus the flux.

Therefore, the action that will not change the magnetic flux through the loop is:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop.

Answer:

d. 37°C

Explanation:

When Equilibrium is reached,

Heat lost = heat gain.

Heat lost by the hot metal = c₁m₁(T₁-T₂)............. equation 1

Where c₁ = specific heat capacity of the metal = 0.25 cal/g⋅°C, m₁ = mass of the metal = 250 g, T₁ = initial Temperature of the metal = 70°C, T₂ final temperature of the metal .

Substituting this values into equation 1,

Heat lost by the metal = 62.5(70-T₂)

Also,

Heat gain by the water = c₂m₂(T₂-T₁)................. equation 2

c₂ = 1.00 cal/g⋅°C., m₂ = 75 g, T₁ = 20°C

Substituting this values into equation 2,

Heat gain by water = 1× 75 (T₂ - 20)

Heat gained by water = 75(T₂ - 20)

Also,

Heat gained by the calorimeter = c₃m₃(T₂-T₁)............. equation 3

Where c₃ = 0.10 cal/ g⋅°C, m =500 g, T₁ =20°C.

Substituting this values into equation 3

Heat gained by the calorimeter = 0.1 × 500(T₂ - 20)

Heat gained by the calorimeter = 50(T₂ - 20)

Heat lost by the metal = heat gained by water + heat gained by the calorimeter.

62.5(70-T₂) = 50(T₂ - 20) + 75(T₂ - 20)

4375 - 62.5T₂ = 50T₂ - 1000 + 75T₂ - 1500

Collecting like terms,

-62.5T₂ - 50T₂ - 75T₂ =  - 1000 - 1500 - 4375

  -187.5T₂ = -6875

Dividing both side by the coefficient of T₂

-187.5T₂ / -187.5 = -6875 /-187.5

 T₂ = 36.666°C

T₂ ≈ 37°C

The final temperature = 37°C

Answer:

The ice is cooler than the table.

Explanation:

The difference in temperature is a necessary condition for the transfer of heat between two bodies.

Here if the heat is being transferred from the table to an ice block this means that the temperature of the table is greater than the ice which drives the heat energy from the table to the ice block and the heat will continue to flow until they both attain a common temperature.

However,

Final answer:

Heat flow from a table to a block of ice enables the ice to undergo a phase change from solid to liquid, known as the latent heat of fusion, without a change in temperature .

Explanation:

If heat is flowing from a table to a block of ice moving across the table, which of the following must be true? The correct answer from the choices provided is that the ice is changing phase. Heat flow from the table to the ice suggests that the ice is absorbing energy. When ice absorbs heat, it undergoes a phase change, where the ice melts and becomes liquid water without any change in temperature. This process is known as the latent heat of fusion, where heat is transferred to cause a phase change without altering the temperature of the system. Therefore, while the table may be rough and friction may exist, and the ice might indeed be cooler than the table, the guaranteed phenomenon occurring is a phase change from solid to liquid as the ice absorbs heat.

Answer:

[tex]|\vec{F}| =48.60\ N[/tex]

Explanation:

given,

Current in the power line = I = 5500 A

earth's magnetic field = 60.0 μT

inclination downward = 67°

Length = 160 m

magnetic force = ?

[tex]\vec{F} = I (\vec{L}\times \vec{B})[/tex]

[tex]|\vec{F}| = I |(\vec{L}\times \vec{B})|[/tex]

[tex]|\vec{F}| =I LB sin \theta [/tex]

[tex]|\vec{F}| = 5500 \times 160 \times 60 \times 10^{-6}\times sin 67^0[/tex]

[tex]|\vec{F}| =48.60\ N[/tex]

According to the right hand rule the direction of the force is perpendicular to the plane of the length and the magnetic field so, it is to west.

Answer:

If the period of the wave is increased by the factor of 2.70, the wavelength of the wave is also increased by a factor of 2.70. So,

[tex]\lambda_2 = 235\times 2.70 = 634.5 ~nm[/tex]

The magnetic field component can be written as

[tex]\vec{B} = \frac{E_{max}}{c}e^{i(\vec{k}\vec{z}-\omega t)}\^{z}[/tex]

The magnetic field is in the z-direction, because the E-field is directed towards +y and the wave is propagating in the +x-direction. The right-hand rule gives us the direction of the B-field.

[tex]\vec{E} \times \vec{B} = \vec{S}[/tex]

S is the Poynting vector which gives us the propagation of the wave.

We will use the following relationships

[tex]k = 2\pi / \lambda\\f = \omega / 2\pi\\c = \lambda f = \lambda \omega / 2\pi\\\omega = 2\pi c/\lambda[/tex]

[tex]\vec{B} = \frac{7.7\times 10^{-3}}{3\times 10^8}e^{(\frac{2\pi}{3\times 10^8}z - \frac{2\pi\times 3\times 10^8}{634.5})}\\\vec{B} = 2.56\times10^{-11} e^{(2.09\times10^{-8}z - 2.96\times10^{6}t)}\^{z}[/tex]

To find the resulting magnetic field equation of an electromagnetic wave after the period is increased, we use relationships between the wave's wavelength, frequency, period, and electric and magnetic fields, applying the changes to the wave's properties based on Maxwell's theory of electromagnetism.

The question is asking for the equation of the resulting magnetic field component of an electromagnetic wave after its period has been increased by a factor of 2.70. Given the initial wavelength λ of 235 μm (micrometers) and the maximum electric field Emax in the +y direction of 7.70×10⁻³ V/m, we can find the initial frequency ν using the relationship ν = c / λ, where c is the speed of light. Since the period T is the reciprocal of the frequency (ν = 1/T), when the period is increased by a factor of 2.70, the frequency decreases by the same factor. The maximum magnetic field Bmax is related to the maximum electric field Emax by Bmax = Emax / c. The resulting magnetic field B can be described by the function B(x, t) = Bmax sin(kx - ωt + φ), where k is the wave number, ω is the angular frequency (2πν), and φ is the phase constant.

Answer:C

Explanation:

There are two types of cells in the eyes rod and cone cells. Rod cells Provide vision during the night or dim light also called of scotopic vision whereas cone cells provide vision during day time or at bright light also called photopic vision.  

Rod cells do not support the color vision that is why it is difficult to differentiate between colors.

Also, the amount of light entering the eyes is low as result cone cells are unable to stimulate.

It is hard to see color at night because cones, which are responsible for color vision, need more light to be stimulated than rods, which allow us to see in low light but only in grayscale.

The best explanation for why it is difficult to discriminate the color of an object at night is option C. At night, the amount of light entering the eye is insufficient to stimulate the cone cells but is sufficient to stimulate the rod cells. The rods are highly sensitive to light, allowing us to see in low light conditions but do not provide color information. Cones require brighter light to function and are responsible for our color vision. Since cones do not react to low-intensity light, our vision at night is predominantly in shades of gray, serviced by the activity of rods.

Answer:

a. [tex]m=\frac{2f_{s}tan\theta}{g}[/tex]

b. m=5.99kg

Explanation:

a.

In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).

In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any  movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.

So we start with the sum of forces about y, so we get:

[tex]\sum F_{y}=0[/tex]

N-W=0

N=W

N=mg

Next we can do the sum of forces about x, so we get:

[tex]\sum F_{x}=0[/tex]

which yields:

[tex]f_{s}-F_{W}=0[/tex]

so:

[tex]f_{s}=F_{W}[/tex]

Next the torque about point B, so we get:

[tex]\sum M_{B}=0[/tex]

so:

[tex]f_{s} L sin\theta - NLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:

[tex]f_{s} L sin\theta - WLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

We can now combine like terms, so we get:

[tex]f_{s} L sin\theta -\frac{WL}{2}cos\theta = 0[/tex]

we know that W=mg, so we can substitute that into our equation, so we get:

[tex]f_{s} L sin\theta -\frac{mgL}{2}cos\theta = 0[/tex]

which can now be solved for the mass m:

[tex]\frac{mgL}{2}cos\theta = f_{s} L sin\theta[/tex]

If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:

[tex]\frac{mg}{2}cos\theta = f_{s}sin\theta[/tex]

we can now divide both sides of the equation into g so we get:

[tex]\frac{m}{2}cos\theta = \frac{f_{s}sin\theta}{g}[/tex]

next we can divide both sides of the equation into cos θ so we get:

[tex]\frac{m}{2}= \frac{f_{s}sin\theta}{g cos\theta}[/tex]

and finally we can multiply both sides of the equation by 2 so we get:

[tex]m=\frac{2f_{s}sin\theta}{g cos\theta}[/tex]

we know that:

[tex]tan \theta=\frac{sin \theta}{cos \theta}[/tex]

so we can simplify the equation a little more, so we get:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

b.

So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:

θ=32° and [tex]f_{s}=47N[/tex]

we also know that [tex]g=9.8m/s^{2}[/tex]

so we can use our equation now:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

so we get:

[tex]m=\frac{2(47N)tan (32^{o})}{9.8m/s^{2}}[/tex]

which yields:

m=5.99kg

Answer:

a) m = 2fs(tanθ)/g

b) m = 5.99 kg

Explanation:

a) Expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.

Data

m₁ : mass of the lader

g: acceleration due to gravity

L : ladder length

θ  : angle that makes  the  ladder  with the floor

µ = 0 : coefficient of friction between the ladder and the wall

fs : friction force between the ladder and the floor

Forces acting on the ladder

W =m*g : Weight of the ladder (vertical downward) , m: mass of the lader 

FN :Normal force that the floor exerts on the ladder (vertical upward) (point A)

fs : friction force that the floor exerts on the ladder (horizontal to the left) (point A)

N : Forces that the wall exerts on the ladder (horizontal to the right)

Equilibrium  of the forces in X

∑Fx=0

N -fs = 0

N = fs

The equilibrium equation of the moments at the point contact point of the ladder with the floor:

∑MA = 0  

MA = F*d  

Where:  

∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)  

MA : moment in the point A ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point A ( m )

Calculation of the distances of the forces at the point A

d₁ = (L/2)*cosθ : Distance from W to the point A

d₂ = L*sinθ : Distance from N to the point A

Equilibrium of the moments at the point A

∑MA = 0  

N(d₂)-W(d₁) = 0

W( (L/2)*cosθ)= N(L*sinθ )

mg( (L/2)*cosθ)= fs(L*sinθ )

We divided by L both sides of the equation

mg (cosθ/2) =fs(sinθ)

m=2fs(sinθ)/ g( cosθ)

m = 2fs(tanθ)/ g

b) If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?

m = 2fs(tanθ)/ g

m = 2(47)(tan32°)/(9,8)

m = 5.99 kg

Answer:

Correct answer is (b) Revitalization movements always fail because they require too much change to be tolerated.

Example of a revitalization movement is the Ghost Dance that swept through western Native American cultures from 1870-1890.

Explanation:

Another example of Revitalization movements are mostly associated with religion. They often occur in disorganized societies due to warfare, revolution but not necessarily an animistic societies.

The incorrect statement is that 'Revitalization movements always fail because they require too much change to be tolerated.' Revitalization movements have led to significant societal changes and have even given rise to major religions like Christianity, Judaism, and Islam. The correct option is b.

The statement 'Revitalization movements always fail because they require too much change to be tolerated' is incorrect. Revitalization movements are deliberate, organized, conscious efforts by members of a society to construct a more satisfying culture. While it's true that they can face intense resistance because they often push for significant changes, it's not true that they always fail. Many have had long-lasting impacts. For example, the Protestant Reformation was a revitalization movement that sought to reform the Roman Catholic Church and led to the creation of Protestant churches. It instigated enormous change in society, and it certainly didn't fail.

All major religions, including Judaism, Christianity, and Islam, did indeed begin as revitalization movements. These movements were adapted to the changing needs of society, and they have resulted in lasting religions that are still prevalent today. Hence revitalization movements can have profound and lasting impacts on society, institutions and cultures, far from always being impractical or bound to fail.

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To develop this problem we will proceed to convert all units previously given to the international system for which we have to:

[tex]140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N[/tex]

[tex]120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N[/tex]

[tex]170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N[/tex]

PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:

[tex]F = 756 - 622.3[/tex]

[tex]F = 133.7N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

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

[tex]a = \frac{133.7}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

PART B) For the maximum magnitude of the acceleration downward we have that:

[tex]F = 622.3 - 533[/tex]

[tex]F = 89.3N[/tex]

Through the Newtonian relationship of the Force we have to:

[tex]F= ma[/tex]

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

[tex]a = \frac{89.3}{63.5}[/tex]

[tex]a = 2.1m/s^2[/tex]

[tex]a = 1.04 m/s^2[/tex]

To solve this problem we will use the concept of the Doppler effect applied to the speed of blood, the speed of sound in the blood and the original frequency. This relationship will also be extrapolated to the frequency given by the detector and measured the change in frequencies through the beat frequency. So:

[tex]f_{blood} = f (1-\frac{v_{blood}}{v_{snd}})[/tex]

Where

[tex]f_{blood}[/tex] = Frequency of the blood flow

f = Frequency of the original signal

[tex]v_{blood}[/tex] = Speed of the blood flow

[tex]v_{snd}[/tex] = Speed of sound in blood

[tex]f''_{detector} = \frac{f_{blood}}{(1+\frac{v_{blood}}{v_{snd}})}[/tex]

[tex]f''_{detector} = f (\frac{(1-\frac{v_{blood}}{v_{snd}})}{(1+\frac{v_{blood}}{v_{snd}})})[/tex]

[tex]f''_{detector} = f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

Now calculating the beat frequency is

[tex]\Delta f = f-f''_{detector}[/tex]

Replacing this latest value we have that,

[tex]\Delta f = f-f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

[tex]\Delta f = f \frac{2v_{blood}}{v_{snd}+v_{blood}}[/tex]

Replacing we have,

[tex]\Delta f = (3.5*10^6)(\frac{2*(3*10^{-2})}{1.54*10^3+3*10^{-2}})[/tex]

[tex]\Delta f = 136.36Hz[/tex]

Therefore the beat frequency is 136.36Hz

Using the beat frequency relation, the expected beat frequency observed at the flow meter would be 136.36 Hz

Given the Parameters :

The expected beat frequency observed can be calculated uisng the relation :

Substituting the values into the formula :

[tex] \delta F = 3.5 \times 10^{3} \frac{2 \times 0.03}{(1540 + 0.03}[/tex]

[tex] \delta F = 3.5 \times 10^{3} \frac{0.06}{(1540.03}[/tex]

[tex] \delta F = 136.36 Hz [/tex]

Therefore, the expected beat frequency observed at the flow meter will be 136.36 Hz

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