Suppose we have no idea what the voltages of our batteries are. What must these voltages be if a 100 amp current ?ows uniformly across all three loops of the circuit above?

Answer:

It is constant.

Explanation:

As we know that the electric field is a change in the electric potential so,

E = -ΔV

E = 0 ∵ V = constant

As we know that the electric field is zero. Which means that the electric potential is constant and it's not changing which results in the zero electric field.

In a region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero.

In a certain region of space where the electric field is zero, it does not necessarily mean that the electric potential in that region is zero. The electric potential can still have a non-zero value even if the electric field is zero. This is because electric potential is determined by the distribution of charges and their distances from the region in consideration. While the electric field measures the force experienced by a charge, the electric potential is related to the work required to move a charge in a region with an electric field.

Therefore, the correct answer is (e) None of those answers is necessarily true.

Answer:

1.29649

488.08706 nm

[tex]6.14644\times 10^{14}\ Hz[/tex]

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

[tex]\lambda_0[/tex] = 632.8 nm

From Snell's law we have the relation

[tex]n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649[/tex]

Refractive index of the solution is 1.29649

Wavelength is given by

[tex]\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm[/tex]

The wavelength of the solution is 488.08706 nm

Frequency is given by

[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz[/tex]

The frequency is [tex]6.14644\times 10^{14}\ Hz[/tex]

[tex]v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s[/tex]

The speed in the solution is 231715700.28346 m/s

The index of refraction can be found using Snell's law. The wavelength of red light within the solution can be calculated from the index and its frequency remains constant. The speed of light in the solution is the product of its frequency and the calculated wavelength.

Solution to the Laser Beam Question

To answer the student's question regarding refraction of a laser beam in a corn syrup solution, we can use Snell's law, which relates the incident angle, refracted angle, and indices of refraction of the two media. We can also calculate the wavelength, frequency, and speed of light within the solution using the principles of wave optics.

(a) According to Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n2 is the index of refraction of the corn syrup solution, θ1 is the incident angle, 90° - 33.0° = 57.0°, and θ2 is the refracted angle, 90° - 24.84° = 65.16°. Since the laser light is passing from air (n1=1) to the solution, we have 1 * sin(57.0°) = n2 * sin(65.16°). Solving for n2 gives us the index of refraction of the syrup solution.

(b) The wavelength of light in the solution is given by λ' = λ0/n, where λ0 is the wavelength in a vacuum, and n is the index of refraction. Inserting the red light's wavelength (632.8 nm) in vacuum and the obtained index of refraction will yield the wavelength in the solution.

(c) The frequency of light does not change when it enters another medium, so we use the vacuum frequency calculated by f = c/λ0, and c is the speed of light in a vacuum.

(d) The speed of light in the solution, v, can be found using the equation v = f * λ', which uses the frequency found in part (c) and the wavelength in the solution from part (b).

Part of the problem involves not only answering the question with the correct degree of accuracy, but also making approximations to the correct answer. For example, for the first case we know that the speed is equivalent to the distance traveled in a given time. Therefore it would be defined as

]1)  [tex]V = \frac{x}{t}[/tex]

x = Displacement

t = Time

The displacement value is 100km and the time value is 5 hours. Therefore the speed value would be

[tex]V = \frac{100km}{5h} = 20km/h[/tex]

We know that the answer margin is within 5% of the value, then 5% of 20 would be 1.   That is, we have a margin of error of '1km / h' to answer the question. Any value that falls within that range can be added or subtracted from the response and the response will be valid. Values included within this value would be

[tex]V_1 = 20km/h +1km/h = 21km/h[/tex]

[tex]V_2 = 20km/h - 1km/h = 19km/h[/tex]

[tex]V_3 = 20km/h + 0.5km/h = 20.5km/h[/tex]

[tex]V_4 = 20km/h -0.5km/h = 19.5km/h[/tex]

[tex]V_5 = 20km/h +0.01km/h = 20.01km/h[/tex]

2) For the second case the margin of tolerance for the response is 5%, so if we multiply the given value we would have a response of.

[tex]x = 6.5*2 = 13[/tex]

5% of 13 is 0.65. Therefore, any value that falls within that range will be a correct answer. The value could then be

[tex]x_1 = 13+0.65 = 13.65[/tex]

[tex]x_2 = 13-0.65 =12.35[/tex]

Incorrect values will be

[tex]x_3 = 13+1 = 14[/tex]

[tex]x_4 = 13-1=12[/tex]

Answer:

v1 = 377.98 m/s

Explanation:

m = 5 Kg

v0 = 176 m/s

v0x = v0*Cos 32° = 176 m/s*Cos 32° = 149.256 m/s

m1 = 2 Kg

m2 = 3 Kg

t = 4.1 s

g = 9.81 m/s²

Before  the explosion

pix = m*v0x = 5 Kg*149.256 m/s = 746.282 Kgm/s

piy = 0

After the explosion

pfx = m1*v1x

knowing that pix = pfx

we have

746.282 = 2*v1x

v1x = 373.14 m/s

v2y = g*t

pfy = m1*v1y + m2*v2y

pfy = 2*v1y + 3*(9.81*4.1)

pfy = 2*v1y + 120.663

knowing that piy = pfy = 0

we have

0 = 2*v1y + 120.663

v1y = 60.33 m/s

Finally we apply

v1 = √(v1x² + v1y²)

v1 = √(373.14² + 60.33²)

v1 = 377.98 m/s

Final answer:

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we use the principle of conservation of momentum. By setting the momentum before the explosion equal to the momentum after the explosion, we can solve for the velocity of the 2 kg fragment. The magnitude of the velocity is approximately 56.8 m/s.

Explanation:

Projectile Motion

To find the magnitude of the velocity of the 2 kg fragment immediately after the explosion, we need to use the principle of conservation of momentum. Since there are no external forces acting on the system, the total momentum before the explosion is equal to the total momentum after the explosion.

Before the explosion, the momentum of the projectile can be calculated using the formula:

momentum = mass * velocity

After the explosion, the momentum of the 2 kg fragment can be calculated as:

momentum = mass * velocity

Setting the two equations equal to each other, we can solve for the velocity of the 2 kg fragment.

Plugging in the given values:

Initial velocity of the projectile = 176 m/s
Initial angle = 32°
Mass of the 2 kg fragment = 2 kg
Mass of the 3 kg fragment = 3 kg
Time after explosion = 4.1 s
Acceleration due to gravity = 9.81 m/s^2

Solving the equation, we find that the magnitude of the velocity of the 2 kg fragment immediately after the explosion is approximately 56.8 m/s.

Answer:

D = 3.45 mile

Explanation:

given,

time taken by the sound wave, t = 16.2 s

speed of sound, v = 343 m/s

speed of the light = 3 x 10⁸ m/s

distance where lightning stroke

 we know,

 distance = speed x time

  D = 343 x 16.2

  D = 5556.6 m

 1 m = 0.000621371 mile

5556.6 m = 5556.6 x 0.0006214

D = 3.45 mile

the distance lightning strike is 3.45 mile away from the observer.

To calculate the distance from a lightning stroke: use the time delay between seeing the lightning and hearing the thunder, and multiply it by the speed of sound (343 m/s). In this case, the lightning is approximately 5562.6 meters or about 3.46 miles away.

Your distance from a lightning stroke can be calculated by knowing the speed of sound and the time it takes for the sound from the lightning stroke to reach you. You said you hear the thunder sound 16.2 seconds after seeing the lightning. The speed of sound in air is 343 m/s. Therefore, the distance can be calculated using the formula: distance = speed * time.

In this case, distance = 343 m/s * 16.2s = 5562.6 meters or about 5.56 kilometers.

When this is converted to miles (since 1 mile is about 1609.34 meters), it is approximately 3.46 miles away. Therefore, based on the time difference between the flash and the thunder, you are about 5562.6 meters or 3.46 miles away from the lightning strike.

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Answer:

The calculated vectors are:

[tex]\vec{A}-\vec{B}=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=(-3,-2,-11)[/tex]

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

[tex]\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)[/tex] (1)

[tex]\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)[/tex] (2)

[tex]\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)[/tex] (3)

The operations in these cases are:

[tex]\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)[/tex]

[tex]\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)[/tex]

[tex]-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)[/tex]

[tex]3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)[/tex]

Answer:

Part A: (3, -5, -4)

Part B: (-5, 4, 0)

Part C: (-6, 5, 3)

Part D: (-3, -2, -11)

Part E: (17, -12, -6)

Explanation:

This problem involves addition and subtraction of vectors. This can be done by adding and subtracting the respective components of each vector as the case may be.

The full descriptive solution can be found in the attachment below.

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge [tex]q=5\sin4\pi t\ mC[/tex]

Voltage [tex]v=3\cos4\pi t\ V[/tex]

Time t = 0.3 sec

We need to calculate the current

Using formula of current

[tex]i(t)=\dfrac{dq}{dt}[/tex]

Put the value of charge

[tex]i(t)=\dfrac{d}{dt}(5\sin4\pi t)[/tex]

[tex]i(t)=5\times4\pi\cos4\pi t[/tex]

[tex]i(t)=20\pi\cos4\pi t[/tex]

(a).We need to calculate the power delivered to the element

Using formula of power

[tex]p(t)=v(t)\times i(t)[/tex]

Put the value into the formula

[tex]p(t)=3\cos4\pi t\times20\pi\cos4\pi t[/tex]

[tex]p(t)=60\pi\times10^{-3}\cos^2(4\pi t)[/tex]

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})[/tex]

Put the value of t

[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})[/tex]

[tex]p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)[/tex]

[tex]p(t)=187.68\ mW[/tex]

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

[tex]E(t)=\int_{0}^{t}{p(t)dt}[/tex]

Put the value into the formula

[tex]E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}[/tex]

[tex]E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}[/tex]

[tex]E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}[/tex]

[tex]E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)[/tex]

[tex]E(t)=57.52\ mJ[/tex]

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

To solve the problem, the power at a certain moment (t=0.3s) is calculated by substitifying the values of voltage and charge into the power formula. The energy delivered to the element can be calculated by integrating the power function over the period from 0 to 0.6 s.

The subject of this question is related to electrical power and energy in a circuit, and it requires a knowledge of trigonometry. The instantaneous power in an electrical circuit is given by the product of charge (q) and voltage (v). Hence we can calculate the power at t = 0.3s by substituting the given values into the power formula.

Power, p = qv = (5 sin 4πt) * (3 cos 4πt) = 15 sin 4πt cos 4πt. At t = 0.3s, power p = 15 sin 4π * 0.3 cos 4π * 0.3 = 15 sin 1.2π cos 1.2π.

For part (b), the energy delivered to the element between 0 and 0.6 s can be obtained from the integral of the power function over the given time interval, which requires integration skills in calculus.

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Answer:

To make it easier to Understand, consider the circle to be in the usual 3 dimensional x-z plane (the "horizontal plane") and the point of measurement C to be at a distance p (instead of x to avoid confusion with the x-axis) on the y-axis (the "vertical direction").

The answer for a and b is the same.  This is because a the horizontal component of a charged portion of a uniform ring is canceled by the opposite portion of the ring since they are placed at an equal distance from the position in which the electric field is being measured or applied, just as are the opposing point charges described in part a.

Explanation:

A.)  Using Coulomb's law of point charges, each charge on the circle would exert a field Ec at C given by:

(1)  Ec   =   Ke * (Q / n) / d²

where:

Ke is Coulomb's constant,

Q / n  is the magnitude of the charge, and

d   =   the distance between the charge and the point of measurement C, with   d²    =    a²  +  c²  

Since the charges are in a circle in the x-z  plane, all force components in both the x- and the z-directions are canceled by symmetry; the vertical force (that in the y-direction) is the only component that does not cancel.  

Therefore the resultant vector Ecy points up (+y-direction) and has a magnitude of:

(2)  Ecy   =   Eq  *  sin(theta)

=  (Ke * (Q / n) / d²)  *  (c  /  d)

Then, summing the forces from all the charges, the magnitude of the total electric field is given by:

(3)  Ey   =   n * Ecy

=   n * [ (Ke * (Q / n) / d²)  *  (c  /  d) ]

=   c * Ke * Q  / d^3

B.) The equation is the same. This is because both the x- and z-components (the two planar components) of a charged portion of a uniform ring are canceled by the opposite portion of the ring,  since they lie at an equal distance but opposite direction from c.  

This is the same way the opposing point charges described in part A behave.

In both scenarios described, because of symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.

The magnitude of the electric field in the scenario described is given by Coulomb's law, which states that the electric field E due to a charge Q at a distance r from a point is given by E = ke*(Q/r^2), where ke is Coulomb's constant.

In this case, the magnitude of each charge is Q/n and we have n charges symmetrically distributed along the circle. Because we are calculating the electric field at a point on the line passing through the center of the circle and perpendicular to the plane of the circle, each pair of charges on the circle will produce an electric field that is symmetric and thus its contribution will be canceled out by another pair's contribution. The final resultant electric field will hence be zero.

(b) Now let's consider a ring of radius a that carries a uniformly distributed positive total charge Q. Because the charge is symmetrically distributed, the electric field at any point on the line passing through the center and perpendicular to the plane of the ring will be sum of electric fields due to all small charged particles which make up this ring, and due to symmetry of the distribution, it will add up to zero.

Therefore, the results in part (a) and (b) are identical because in both cases, due to symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.

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Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

The change in the sound intensity level in dB is -13.1 dB.

The given parameters;

The relationship between intensity of sound and distance is calculated as follows;

[tex]I = \frac{k}{r^2} \\\\I_1r_1^2 = I_2r_2^2\\\\I_2 = \frac{I_1 r_1^2 }{r_2^2} \\\\I_2 = \frac{I_1 r_1^2}{(4.49r_1)^2} \\\\I_2 = \frac{I_1r_1^2}{20.16r_1^2} \\\\I_2 = \frac{I_1}{20.16} \\\\\frac{I_2}{I_1} = \frac{1}{20.16}[/tex]

The change in sound intensity in dB is calculated as follows;

[tex]\Delta \beta = 10 \ log[\frac{I_2}{I_1} ]\\\\\Delta \beta = 10 \times log [\frac{1}{20.16} ]\\\\\Delta \beta = -13.1 \ dB[/tex]

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To develop this problem we will start using the concept of maximum speed for this type of systems. The maximum velocity can be described as the product between the Amplitude and the Angular velocity. At the same time, said angular velocity can be found through the relationship between linear and "angular wavenumber" velocity. The Angular wavenumber is a wave number defined as the number of radians per unit distance. Finally with the value of the angular velocity found we will proceed to find the maximum speed.

The maximum speed is given by

[tex]v_{max} = A\omega[/tex]

Here,

A = Amplitude

[tex]\omega[/tex]= Angular velocity

The angular velocity can be described as the number of radians per unit distance

[tex]\omega = vk[/tex]

[tex]\omega = v (\frac{2\pi}{\lambda})[/tex]

[tex]\omega = 112(\frac{2\pi}{12.16})[/tex]

[tex]\omega =57.8714rad/s[/tex]

Then,

[tex]v_{max} = 0.12 *57.8714[/tex]

[tex]v_{max} = 6.94m/s[/tex]

Therefore the maximum speed a point on the medium moves as this wave passes is 6.94m/s

When you slide a coin across the floor, it slows down and eventually stops due to the force of friction. The friction converts the coin's kinetic energy into thermal energy, resulting in an increase in temperature.

When you slide a coin across the floor, it eventually slows down and stops due to the force of friction acting on it. Friction is a force that opposes the motion of objects in contact, and it causes the coin to lose kinetic energy, slowing it down. As the coin slows down, its kinetic energy is converted into thermal energy, increasing the temperature of the coin and the surface it slides on. This is why a sensitive thermometer shows an increase in temperature when the coin slides.

Answer:

4.662 slugs

Explanation:

Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.

If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is

m = W / g = 150 / 32.174 = 4.662 slugs

which is also your mass on the moon.

Answer:

v = 5.05m/s

Explanation:

H = 1.3m

initial velocity = 0

final velocity = v = ?

g =9.8 m/s^2

we apply the conservation of energy; all potential energy is comletely converting to kinetic energy

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

mass is same; it cancels out

[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]

v = 5.05m/s

Answer:

the final linear velocity is 3.56931 m/s

Explanation:

the solution is in the attached Word file

Answer:

9.6609 rad/s

10.143945 m/s

Explanation:

I = Moment of inertia = 3.75 kgm²

K = Kinetic energy = 175 J

r = Radius = 1.05 m

Kinetic energy is given by

[tex]K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s[/tex]

The angular velocity of the leg is 9.6609 rad/s

Velocity is given by

[tex]v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s[/tex]

The velocity of the tip of the punters shoe is 10.143945 m/s

Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are  in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

To find the equivalent resistance Req between terminals a and b, we can follow a series of steps to simplify the circuit. The final equivalent resistance can be calculated by adding the resistors in series. With terminals c and d open, Req is 12.22 ohms, and with terminals c and d shorted, Req is 22.00 ohms.

To find the equivalent resistance between terminals a and b, we need to consider the circuit shown in Figure 10.15. To simplify the circuit, we can apply a series of steps to reduce it to a single equivalent resistance. Step 1 involves reducing resistors R3 and R4 in series. Step 2 involves reducing resistors R2 and the equivalent resistance R34 in parallel. Finally, Step 3 involves reducing resistor R1 and the equivalent resistance R234 in series. The final equivalent resistance Req can be calculated by adding the resistors in series.

Following these steps, the equivalent resistance Req between terminals a and b with terminals c and d open is 12.22 ohms. With terminals c and d shorted together, the equivalent resistance Req is 22.00 ohms.

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Answer:

The pressure is expected to decrease because as we move up in the atmosphere the is lesser mass of atmosphere weight above us.

Explanation:

The atmospheric pressure is the the force exerted by the layers of the atmosphere above a point. Being a fluid and due to the gravity of earth there is a pressure due to the weight of the atmospheric columns above us.

So, at a height h above the earth surface in the atmosphere the pressure will decrease and this can be calculated as:

[tex]P_h=101325\times (1-2.25577\times 10^{-5}\times h)^{5.25588}[/tex]

putting h=9.6 meter

[tex]P_h=101209.7268\ Pa[/tex]

At an elevation of 9.6 km, the pressure is approximately 254 millibars.

The pressure is inversely proportional to elevation, higher the elevation lower will be the pressure while on the other hand, lower the elevation higher will be the pressure. At an elevation of 10 km pressure drops to 265 millibars then the elevation of 9.6 km, the pressure is approximately 254 millibars.

The atmospheric pressure at sea level is about 1,014 millibars which is equal to the pressure of 14.7 pounds/inch2 so we can conclude that at the elevation of 9.6 km, the pressure is approximately 254 millibars.

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Final answer:

The question pertains to the physics discipline, dealing with calculating the tension in supporting cables using principles of static equilibrium and considering the load's weight and the angles of attachment.

Explanation:

The question involves determining the tension in cables that support a structure or an object. This type of problem is common in physics, specifically in the area of mechanics, and it involves understanding forces and how they are distributed within a system. When multiple cables support a load, tensions in each cable can be found using principles of static equilibrium, where the sum of forces in each direction (horizontal and vertical) equals zero. In addition to the weight of the supported object, angles at which cables are attached can play a crucial role in determining individual tensions.

Answer:

Explanation:

Given

Charge on metal sphere [tex]Q=12\mu C[/tex]

no of electrons [tex]n=5.9\times 10^{13}[/tex]

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

Charge by Possessed by Electrons [tex]Q_2=-1.6\times 10^{19}\times 5.9\times 10^{13}[/tex]

[tex]Q_2=-9.44\mu C[/tex]

Net Charge on Sphere [tex]Q_{net}=Q+Q_2[/tex]

[tex]Q_{net}=12-9.44=2.56\mu C[/tex]

Answer:

D

Explanation:

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction.

When a bus suddenly stops, a passenger tends to keep moving forward due to inertia (Newton's first law). The force causing you to fall forward in this instance is the insufficient static friction between you and the floor of the bus, which isn't enough to counteract your forward momentum.

The force acting on you that causes you to fall forward when the bus comes to an immediate stop is B) the force due to static friction between you and the floor of the bus. This is because of Newton's first law, also known as the law of inertia, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless acted upon by an external force.

While you're standing in the moving bus, both you and the bus are moving forward. When the bus suddenly stops, your body tends to keep moving forward due to inertia, causing you to fall forward. Here, the static friction between you and the floor of the bus isn't enough to counteract your momentum, leading to your forward fall.

It's important to note that other forces such as gravity and normal force are in effect too. Gravity pulls you downward, and the bus floor exerts an upward normal force to counterbalance it. These two forces are balanced and don't contribute to your forward movement. The force causing the unbalanced motion (falling forward) is the lack of sufficient friction to oppose your inertia.

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Answer:

(a). 12 plants

(b). 3171 $

Explanation:

(a)first convert units of 100 billion kWh/year into Watts(W)

also convert the units of 1000 MW into Watts(W)

1 billion = 10^9

1 year = 365*24 = 8760 hrs

so

100 billion kWh/year = 1[tex]\frac{100*(10^9)*(10^3)}{8760}[/tex]

                                  = [tex]1.142*10^{10}[/tex]W

1000 MW                  = [tex]1000*10^{6} = 10^{9}W[/tex]

no. of plants = [tex]\frac{1.14155*10^{10} }{10^9}[/tex] = 11.4

So 12 plants required        

(b)

savings = unit price*total units

             = [tex]0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )[/tex]

             = 3170.9 =3171 $

Answer:

a) Number of generating plants N = 11.42

That means N > 11

N = 12

b) annual savings S = $1×10^10

S = $10 billion

Explanation:

Given;

Amount of energy to be saved A=100 billion kWh/year

Capacity of each generating plant C= 1000 MW

Rate in dollars of cost of electricity R= $0.10/kWh

The number N of generating plants with capacity C that can supply the Amount A of of energy cam be given as;

N = A/C ......1

And also the Annual savings S in dollars if the rate of electricity cost R is used and amount of energy A is saved is:

S = AR .....2

But we need to derive the value of A in Watts

A = 100 billion kWh/year

There are 8760hours in a year,

A = 1×10^14 ÷ 8760 W

A = 11415525114.1W or 11415525.1141kW

C = 1000MW = 1× 10^9 W

a) Using equation 1,

N = 11415525114.1/(1×10^9)

N = 11.42

That means N > 11

N = 12

b) using equation 2

S = 1×10^11 kWh × $0.10/kWh

S = $1×10^10

S = $10 billion

Answer:

a) the mole fraction of air in the exiting stream is Xa=0.25 (25%) and for methane  Xm=0.75 (75%)

b) the volumetric flow rate is 49.33 L/s

Explanation:

Assuming ideal gas behaviour, then

for air

Pa*Va=Na*R*Ta

for methane

Pm*Vm=Nm*R*Tm

dividing both equations

(Pa/Pm)*(Va/Vm)= (Na/Nm)*(Ta/Tm)

Na/Nm = (Pa/Pm)*(Va/Vm) * (Tm/Ta) = (0.2/0.2)*(20/5)*(300/400) =  1*4*3/4 = 3

Na=3*Nm

therefore the moles of gas of the outflowing stream are (assuming that the methane does not react with the air):

Ng= Na+Nm = 4*Na

the mole fraction of A is

Xa= Na/Ng= Na/(4*Na) = 1/4 (25%)

and

Xm= 1-Xa = 3/4 (75%)

also for the exiting gas

Pg*Vg=Ng*R*Tg  = Na*R*Tg + Nm*R*Tg = Pa*Va * (Tg/Ta) + Pm*Vm * (Tg/Tm)

Vg = Va * (Pa/Pg)*(Tg/Ta) + Vm *(Pm/Pg)* (Tg/Tm)

Vg = 20 L/min * (0.2/0.1)*(370/400) + 5 L/min * (0.2/0.1)*(370/300) = 49.33 L/s

Final answer:

The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.

Explanation:

When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:

F = k * (Q1 * Q2) / r^2

where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

We know that the gravitational force is equal to the weight of the paper:

Fg = m * g

where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).

At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:

F = Fg

k * (Q1 * Q2) / r^2 = m * g

We can solve this equation for the magnitude of the charge Q1 or Q2:

Q1 * Q2 = (m * g * r^2) / k

Substituting the known values, we get:

Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)

Q1 * Q2 = 1.383 x 10^-12 C^2

To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:

Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C

The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.

Answer:

The angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating.

Explanation:

It is given that,

Radius of the wheel, r = 0.1 m

Initial angular velocity of the wheel, [tex]\omega_i=35\ rev/s=219.91\ rad/s[/tex]

Final angular velocity of the wheel, [tex]\omega_f=15\ rev/s=94.24\ rad/s[/tex]

Time, t = 3 s

We need to find the angular acceleration of a point on the wheel. It is given by the rate of change of angular velocity divided by time taken. It is given by :

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]

[tex]\alpha =\dfrac{(94.24-219.91)\ rad/s}{3\ s}[/tex]

[tex]\alpha =-41.89\ rad/s^2[/tex]

So, the angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating. Hence, this is the required solution.

Final answer:

The angular acceleration of a point on the wheel can be found using the formula: angular acceleration = change in angular velocity / time interval. Given the initial and final angular velocities and the time interval, we can calculate the angular acceleration.

Explanation:

The angular acceleration of a point on the wheel can be found using the formula:

angular acceleration = change in angular velocity / time interval

Given that the initial angular velocity is 35 rev/s, the final angular velocity is 15 rev/s, and the time interval is 3.0 s, we can substitute these values into the formula:

angular acceleration = (15 rev/s - 35 rev/s) / 3.0 s

Simplifying the equation, we get:

angular acceleration = -20 rev/s / 3.0 s

Converting rev/s to rad/s, we have:

angular acceleration = -20 rev/s ×(2π rad/rev) / 3.0 s

angular acceleration ≈ -41.89 rad/s²

Therefore, the angular acceleration of a point on the wheel is approximately -41.89 rad/s².

Answer:

Yes it is possible to charge balloon to several thousand of volts and the balloon will also store several Joules of energy.

Explanation:

By rubbing a balloon on one's hair or on different types of clothing, the balloon either gain or loss electron.

If the balloon gains electron it becomes negatively charged, it contains more electron and subsequently charged to several thousand volts.

Also, if the balloon losses electron, it becomes positively charged. In this case it contains more proton, which makes the balloon positively charged to several thousand volts.

However, amount of joules depends on volts produced in the balloon.

Volt = Joules/coulomb,

Joules = volts*coulomb

Because the charge of the particles (electron and proton) are small, amount of joules will always be small than volts.

So, it is possible to charge balloon to several thousand of volts and the balloon will store several Joules of energy

Answer:

i) They are equal . ii) It is thousands of times greater for the electron.

Explanation:

i) By definition, the electric field is the electric force per unit charge. If the field is the same, the force will depend on the value of the charge under the influence of the field.

As the magnitude of  the charge of the electron and the proton are the same, we conclude that the electric force on both must be equal in magnitude.

ii) The acceleration on both particles must meet the Newton´s 2nd Law, so, if the forces are equal in magnitude (neglecting any other external interaction), the acceleration will only depend on the mass of both particles, according this general expression:

a = F/m

As the mass of the electron is approximately two thousands times smaller than the proton´s, it concludes that the acceleration on the electron must be thousands of times greater for the electron.

The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. The magnitudes of their accelerations are equal.

(i) The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. This is because the electron has a much smaller mass compared to the proton, and the electric force depends on the charge and mass of the particle.

(ii) The magnitudes of their accelerations are equal. The acceleration of a particle in an electric field depends only on the magnitude of the electric field and the mass of the particle. Since both the electron and proton experience the same electric field, their accelerations will be the same.

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Answer:

Explanation:

Given

Area of capacitor Plates [tex]A=L\times L[/tex]

distance between plates is d

capacitance C is given by

[tex]C=\frac{\epsilon A}{d}[/tex]

[tex]C=\frac{\epsilon \cdot L^2}{d}[/tex]

Provided V is Voltage

[tex]Charge(Q)=capacitance(C)\times Voltage(V)[/tex]

If L is doubled

Capacitance [tex]C'=\frac{\epsilon \cdot (2L)^2}{d}[/tex]

[tex]C'=4\times \frac{\epsilon \cdot L^2}{d}[/tex]

Electric field is given by

[tex]E=\frac{Q}{\epsilon _0A}[/tex]

[tex]E_i=\frac{Q}{\epsilon _0L^2}---1[/tex]

[tex]E_f=\frac{Q}{\epsilon _0(2L)^2}---2[/tex]

divide 1 and 2 we get

[tex]\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}[/tex]

[tex]\frac{E_f}{E_i}=\frac}{1}{4}[/tex]

Answer: the maximum compression force = 59.47 kip

the minimum compression force = 43.33 kip

Explanation:

In this we required to do force balance in order to get maximum and minimum compression force in the column.

the picture below explains the steps to solve the question with diagrams to ease understanding.

Answer:

W = qV

Explanation:

Let V be the potential difference of the battery, and q be the charge on the electron.

The work done in moving a charge through a potential difference V =?

Work = force * distance

Work = F.r

F = qE

But E = V / r

F = q. V / r

Work ( W ) = (qV/ r ) * r

Work = qV

Therefore the force required to move a charge through a potential difference V = qV

The amount of work needed to move an electron from the positive to the negative terminal of a battery using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery is  1.92 x 10-18 joules.

In order to move an electron from the positive to the negative terminal of a battery, you would need to do work on the electron.

The amount of work, W, can be calculated using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery.

The charge of an electron is approximately 1.6 x 10-19 coulombs.

For example, if the voltage of the battery is 12 volts, the work done on the electron would be:
W = ([tex]1.6[/tex]×[tex]10^{-19}[/tex][tex]C[/tex]) × ([tex]12[/tex] [tex]v[/tex]) = [tex]1.92[/tex]×[tex]10^{-18}[/tex] joules.

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Answer:

d. 4A

Explanation:

Given that

Amplitude = A

We know that the distance cover by a particle before coming the the rest from the mean point in the oscillation motion is known as amplitude.

The distance cover by particle from 1 - 2 = A

The distance cover by particle from 2 - 1 = A

The distance cover by particle from 1 - 3 = A

The distance cover by particle from 3-4 = A

Therefore the total distance cover by particle = A+A+A+A = 4 A

Therefore the total distance = 4 A

That is why the answer will be 4 A.

d. 4A

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]

Part b

[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]

(a) The required average speed of the truck during collision is 11 m/s.

(b) The required time interval for the collision is 0.058 s.

(c)  The required magnitude of the average force exerted by the wall on the truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e) The required approximations are:

The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.

Given data:

The mass of the truck is, m = 2100 kg.

The speed of the truck is, v = 22 m/s.

The distance crumpled by the truck is, d = 0.62 m.

(a)

Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,

[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]

Thus, the required average speed of the truck during collision is 11 m/s.

(b)

Now, apply the first kinematic equation of motion as,

[tex]v' = v + at[/tex]

Here, a is the linear acceleration and t is the time interval for the collision.

Solving as,

[tex]0 = 22 + at\\\\ a = -22/t[/tex]

Now, apply the second kinematic equation as,

[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]

Thus, we can conclude that the required time interval for the collision is 0.058 s.

(c)

The expression for the magnitude of average force exerted by wall on truck is,

[tex]F_{av.} = \dfrac{mv}{t}[/tex]

Solving as,

[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]

Thus, the required magnitude of the average force exerted by the wall on truck is  [tex]7.96 \times 10^{5} \;\rm N[/tex].

(d)

The ratio of force of the truck and the gravitation force on the truck is,

[tex]= \dfrac{F_{av}}{mg}[/tex]

Here, g is the gravitational acceleration.

Solving as,

[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]

Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.

(e)

The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.

This is because:

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