A sled is pulled along a level path through snow by a rope. A 20-lb force acting at an angle of 40° above the horizontal moves the sled 90 ft. Find the work done by the force. (Round your answer to the nearest whole number.)

Answer:

Windmill will complete 0.27 revolutions.

Explanation:

We have equation of motion s = ut + 0.5at²

Initial speed, u = 12.5 rad/s

Time, t = 12 seconds.

Angular acceleration, a = -0.75 rad/s²

Substituting

               s = 12.5 x 12 - 0.5 x 0.75 x 12² = 96 rad

1 revolution = 360 rad

[tex]96rad=\frac{96}{360}=0.27revolution[/tex]

So, windmill will complete 0.27 revolutions.

Answer:

Length of wire=2.16 m

Diameter of wire=0.232 mm

Explanation:

m= mass of copper wire= 0.82 g

R= Resistance of copper wire= 0.87 ohms

D= Density of copper= 8.96 g/cm^3

ρ= Resistivity= 1.7×10^-8 Ωm

[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]

[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]

[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

[tex]V_{ab}[/tex] = i R

[tex]V_{ab}[/tex] = (0.1832) (65)

[tex]V_{ab}[/tex] = 11.91 Volts

(c)

Power dissipated in the resister R is given as

[tex]P_{R}[/tex] = i²R

[tex]P_{R}[/tex] = (0.1832)²(65)

[tex]P_{R}[/tex] = 2.18 Watt

Power dissipated in the internal resistance is given as

[tex]P_{r}[/tex] = i²r

[tex]P_{r}[/tex] = (0.1832)²(0.5)

[tex]P_{r}[/tex] = 0.0168 Watt

To compute the values in the question, one can utilize Ohm’s law and the power dissipation formula. The current in the circuit is 0.183 A, the terminal voltage of the battery is 11.908 V, the power dissipated in the 65.0 Ω resistor is 2.18 W, and the energy wasted in the battery's internal resistance is 0.0167 W.

To calculate the current in the circuit, we can use Ohm's law which states that the current is equal to the voltage divided by the resistance. This gives us: I = emf/(R+r) where I is the current, emf is the electromotive force of the battery, R is the load resistance and r is the internal resistance of the battery.

Substituting the values given in the question, we get: I = 12.0 V/(65.0 Ω + 0.5 Ω) = 0.183 A. Therefore, the current flowing through the circuit is 0.183 Amperes.

Now that we have the current, we can calculate the terminal voltage of the battery. The terminal voltage (Vab) is given by the equation: Vab = emf - Ir. Substituting the values, we get: Vab = 12.0V - (0.183A * 0.5Ω) = 11.908 V. Therefore, the terminal voltage of the battery is 11.908 Volts.

Lastly, we can calculate the power dissipated in the resistor R and in the battery’s internal resistance r using the formula: Power = I²R. For the resistor, power dissipation = (0.183A)² * 65.0Ω = 2.18 Watts, and for the battery's internal resistance, power dissipation = (0.183A)² * 0.5Ω = 0.0167 Watts.

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Answer:

15.7 x 10⁻⁶ A

Explanation:

r = radius of the circular loop = 0.03 m

Area of the loop is given as

A = πr² = (3.14) (0.03)² = 0.002826 m²

R = Resistance of the resistor = 189 Ω

ΔB = Change in magnetic field = 0.25 - 0.35 = - 0.10 T

Δt = time interval = 1 sec

Current is given as

[tex]i = - A\left ( \frac{\Delta B}{\Delta t} \right )[/tex]

[tex]i = \left ( \frac{A}{R} \right )\left ( \frac{\Delta B}{\Delta t} \right )[/tex]

[tex]i = \left ( \frac{-0.002826}{18} \right )\left ( \frac{- 0.10}{1} \right )[/tex]

i = 15.7 x 10⁻⁶ A

Answer:

It is actually in both.

Explanation:

The momentum of an isolated system is conserved for both elastic and inelastic collision. OPtion C is correct

Collision is the process by which two bodies come into contact with the release of energy.

Collision can be elastic or inelastic.

Foe elastic collision, both momentum, and energy is conserved while for inelastic collision only momentum is conserved.

From the explanation above, we can see that momentum of an isolated system is conserved for both elastic and inelastic collision

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Answer:

313.6 m downward

Explanation:

The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.

In fact, we have:

[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]

where

y(t) is the vertical position of the projectile at time t

h is the initial height of the projectile

[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally

t is the time

a = g = -9.8 m/s^2 is the acceleration due to gravity

We can rewrite the equation as

[tex]y(t)-h = \frac{1}{2}gt^2[/tex]

where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we  find

[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]

So the bullet has travelled 313.6 m downward.

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

The resultant force acting on the object is (6.00 i + 15.00 j) N, and the magnitude of the resultant force is 16.16 N.

To find the resultant force acting on a 3.00-kg object with an acceleration given by a = (2.00 i + 5.00 j) m/s2, we use Newton's second law of motion (Force = mass  imes acceleration).

For the magnitude of the resultant force, we calculate it using the Pythagorean theorem:

Answer:

EMF = 108.3 Volts

Explanation:

As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop

So it is given as

[tex]EMF = N\frac{d\phi}{dt}[/tex]

[tex]EMF = N\frac{BA - 0}{\Delta t}[/tex]

now we know that

N = 50 turns

B = 1.5 T

A = 0.325 m^2

[tex]\Delta t = 0.225 s[/tex]

now we have

[tex]EMF = (50)(\frac{1.5\times 0.325}{0.225})[/tex]

[tex]EMF = 108.3 Volts[/tex]

The magnitude of the average induced emf is 110.12 volts.

The magnitude of the average induced emf can be calculated using Faraday's law. The formula is given as:

emf = -N * A * (dB/dt)

Where:

In this case, the number of turns is 50, the area is 0.325 m², and

the rate of change of the magnetic field is (0 T - 1.5 T) / 0.225 s = -6.67 T/s.

Substituting these values into the formula, we get:

emf = -(50)(0.325)(-6.67) = 110.12 V

The magnitude of the average induced emf is 110.12 volts.

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Answer:

B) 206.323

Explanation:

The value of (8 104)2, written with the correct mumber of significant figures is 206.323.

Answer:

B. 206.323

Explanation:

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

Answer:

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

Explanation:

We start from:

[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]

Separating variables:

[tex]-50dT=(T-21)dt[/tex]

[tex]-50\frac{dT}{T-21}=dt[/tex]

Integrating with initial conditions:

[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]

[tex]-50ln(\frac{T-21}{91-21})=t[/tex]

[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]

Isolating T:

[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]

[tex]T=21+70e^{\frac{-t}{50} }[/tex]

You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)

The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.

In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).

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Answer:

105.8 degree from + X axis

Explanation:

V1 = 46 N at 30 degree

V2 = 63 N at 151 degree

Write the vector form

V1 = 46 (Cos 30 i + Sin 30 j) = 39.84 i + 23 j

V2 = 63 (Cos 151 i + Sin 151 j) = - 55 i + 30.54 j

The resultant of V1 and V2 is given by

V = V1 + V2 = 39.84 i + 23 j - 55 i + 30.54 j = - 15.16 i + 53.54 j

The angle made by resultant of V1 and V2 with X axis is

tan∅ = 53.54 / (- 15.16) = - 3.53

∅ = 105.8 degree from + X axis.

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

(a) The emf induced in the loop is 0.0171 V.

(b) The current in the loop is determined from the ratio of induced emf to resistance in the loop.

The emf induced in the loop is calculated by applying Faradays law as follows;

emf = dФ/dt

emf = A dB/dt

where;

emf = 0.09 x 0.19

emf = 0.0171 V

The current in the loop is determined by applying Ohm's law;

I = emf/R

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Answer:

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Explanation:

As we know that electric field due to long cylinder on a cylindrical Gaussian surface must be constant

so on the Gaussian surface we will have

[tex]\int E. dA = \frac{q_{en}}{\epsilon_0}[/tex]

now the electric field is passing normally through curved surface area of the cylinder

so we will have

[tex]E (2\pi rL) = \frac{q_{en}}{\epsilon_0}[/tex]

here enclosed charge in the cylinder is given as

[tex]q_{en} = \lambda L[/tex]

from above equation

[tex]E(2\pi rL) = \frac{\lambda L}{\epsilon_0}[/tex]

[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]

Final answer:

The magnitude of the current that flows through the wire is approximately 16A.

Explanation:

To calculate the magnitude of the current that flows through the wire, we can use the formula for the magnetic field produced by a current-carrying circular loop: B = μ0 * I / (2 * R), where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and R is the radius of the loop.

Given that the radius of the loop is 3.0 mm and the magnetic field strength at the center of the loop is 1.1 mT, we can rearrange the formula to solve for I:

I = (2 * B * R) / μ0

Substituting the given values, we get:

I = (2 * 1.1 x 10^-3 T * 3.0 x 10^-3 m) / (4π x 10^-7 T.m/A)

Simplifying the expression, we find that the magnitude of the current is approximately 16 A. Therefore, the correct answer is B) 16A.

Answer:

0.24

Explanation:

w=vt

then

w1/v1=w2/v2

0.4/0.5=w2/0.3

w2= 0.3*0.4/0.5=0.24//

The angular acceleration of the bar is 0.75 rad/s², and the acceleration of end A is 0.3 m/s² in the opposite direction to end B. These values were determined using the relationships between linear and angular velocities and accelerations.

To determine the angular acceleration of the bar and the acceleration of end A, let's go through the step-by-step process:

Determine the Angular Velocity

Given that the bar has a length of 0.4 meters and the velocity of end B is 0.5 m/s, we start by calculating the angular velocity ω. The formula relating linear velocity v and angular velocity ω is:

[tex]v = \omega * r[/tex]

where r is the length of the bar. Rearranging for ω :

[tex]\omega = v / r = 0.5 m/s / 0.4 m = 1.25 rad/s[/tex]

Determine the Angular Acceleration

Next, we use the given acceleration of end B, which is 0.3 m/s². To find the angular acceleration α, we use the formula:

[tex]a = \alpha * r[/tex]

Rearranging for α:

[tex]\alpha = a / r = 0.3 m/s^2 / 0.4 m = 0.75 rad/s^2[/tex]

Determine the Acceleration of End A

The acceleration of end A, which is located at the opposite end of the bar, can be computed using:

[tex]a_A = \alpha * r[/tex]

Since end A is on the same bar, r here will be the same:

[tex]a_A = 0.75 rad/s^2* 0.4 m = 0.3 m/s^2[/tex]

Therefore, the acceleration of end A is also 0.3 m/s², but in the opposite direction to end B.

By using similar triangles concept and rates of change in Mathematics, we can find the man's shadow length is decreasing at a rate of 0.5 m/s when he is 4 meters away from the building.

The problem you're describing is typically solved using similar triangles concept in Mathematics and rates of change. As the man walks towards the building, his shadow becomes shorter. We can set up a ratio between the man's height and his distance from the building, and the length of the shadow and the wall. When the man is 4 meters away from the building, a right-triangle is formed with the man's height (2m), his distance from the building (4m), and his shadow's length.

Let s gives the length of the shadow, then the similar triangles imply 12/s = 2/4 which gives us s = 6 meters.

To find the rate at which the man's shadow is decreasing, we can differentiate the similar triangles ratio with respect to time to get -12/s^2 ds/dt =-1/2, sub in s = 6 into the equation to find ds/dt = -0.5 m/s.

So the length of the man's shadow on the building is decreasing at a rate of 0.5 m/s when he is 4 m from the building.

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Final answer:

To find how fast the man's shadow is decreasing, we use similar triangles and related rates in calculus. When the man is 4 m from the building, his 2 m tall shadow is shortening at 3.8 m/s, twice his walking speed of 1.9 m/s.

Explanation:

The question involves determining how fast the length of a man's shadow on a building is decreasing when he walks toward the building and away from the spotlight. This requires the application of similar triangles and the concept of related rates in calculus to solve. If we let the distance between the man and the building be x, and the length of the shadow be y, as the man walks towards the building, both x and y are changing with respect to time.

Given that the man is 2 m tall and the spotlight is 12 m away from the building, we can set up a proportion between the heights of the man and his shadow, and their respective distances from the spotlight. When he is 4 m from the building, the similar triangles can be represented as 2/y = (12-x)/(12). Differentiating both sides with respect to time t yields the related rate of change for the length of the shadow y when x is 4 m and the speed of the man is known, which is 1.9 m/s.

By solving this proportion and applying the derivative, we can find that the rate at which the length of his shadow decreases is exactly double the speed at which the man is walking, because in this scenario, the man and the tip of his shadow form a straight line with the light source. Thus, the length of the shadow decreases at a rate of 3.8 m/s when the man is 4 m from the building.

Answer:

A

Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

Explanation:

It is given that,

The thinnest soap film appears black when illuminated with light with a wavelength of 535 nm, [tex]\lambda=5.35\times 10^{-7}\ m[/tex]

Refractive index, [tex]\mu=1.33[/tex]

We need to find the thickness of soap film. The soap film appear black means there is an destructive interference. The condition for destructive interference is given by :

[tex]2t=m\dfrac{\lambda}{\mu}[/tex]

t = thickness of film

m = 0,1,2....

[tex]\mu[/tex] = refractive index

[tex]t=m\dfrac{\lambda}{2\mu}[/tex]

[tex]t=\dfrac{\lambda}{2\mu}[/tex]

For thinnest thickness, m = 1

[tex]t=1\times \dfrac{5.35\times 10^{-7}\ m}{2\times 1.33}[/tex]

[tex]t=2.01\times 10^{-7}\ m[/tex]

Hence, this is the required solution.

Final answer:

The thinnest soap film that appears is approximately 50.47 nm due to destructive interference.

Explanation:

The question is asking for the thinnest soap film that appears black when illuminated with light of a specific wavelength, in this case, 535 nm. The phenomenon described is known as thin film interference, which occurs when light waves reflected off the top and bottom surfaces of a film interfere with each other.

The film appears black at the thinnest point where destructive interference occurs, which means the reflected light waves are out of phase and cancel each other out.

To find the thinnest film thickness that appears black, we can use the formula for destructive interference in thin films, taking into account the phase shift that occurs upon reflection from a medium with a lower index of refraction to a higher one. This formula is:
[tex]2nt = (m + \(\frac{1}{2}\))\(\lambda\),[/tex]where n is the index of refraction, t is the thickness of the film, \(\lambda\) is the wavelength of light in vacuum, and m is an integer representing the order of the interference.

For the thinnest film, we use m = 0.

Therefore, the thinnest soap film thickness t is calculated as:

[tex]2nt = \(\frac{1}{2}\)\(\lambda\)2nt = \(\frac{1}{2}\) \\times 535 nm / 1.33t = \(\frac{535 nm}{(2 \\times 1.33 \\times 2)}\)t = 100.94 nm[/tex]

However, regarding physical film thickness, we should only consider the distance that light travels within the film, which is factored by n to give the optical path length. Thus, the actual thickness would be half of 100.94 nm, yielding a value of approximately 50.47 nm.

Answer:

The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.

Explanation:

Given that,

Angular velocity = 5.40 rad/s

Time t = 0.000 sec

Angular acceleration = 1.50 rad/s^2

(a). We need to calculate the angle at time t = 4.00 s

Using formula for angle

[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t^2[/tex]

Where, [tex]\omega_{0}[/tex]=angular velocity

[tex]\alpha[/tex]=angular acceleration

t = time

Put the value into the formula

[tex]\theta=5.40\times4.00+\dfrac{1}{2}\times1.50\times(4.00)^2[/tex]

[tex]\theta=33.6\ rad[/tex]

[tex]\theta=\dfrac{33.6}{2\pi}\ rad[/tex]

[tex]\theta=5.35\ rev[/tex]

(b). We nee to calculate the angular velocity at 4.00 s

Using formula of angular velocity

[tex]\omega=\omega_{0}+\alpha t[/tex]

[tex]\omega =5.40+1.50\times4.00[/tex]

[tex]\omega=11.4\ rad/s[/tex]

[tex]\omega=\dfrac{11.4}{2\pi}\ rad/s[/tex]

[tex]\omega=1.81\ rev/s[/tex]

Hence, The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.

Answer:

(a) 83.73 rad/s

(b) 251.2 m/s

(c) 3505.4 rad/s^2

(d) 5024 m

Explanation:

R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps

(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s

(b) The relation between linear speed and the angular speed is

V = r w

Here, r = 30 cm = 0.3 m

V = 0.3 x 83.73 = 25.12 m/s

(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2

(d) Time period T = 2 x 3.14 / w

T = 2 x 3.14 / 83.73 = 0.075 sec

In 0.075 second, angle turn = 360 degree

In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree

In 360 degree, the distance traveled = 2 x pi x R

In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360

                                                                     = 5024 m

Answer:

24.2 degree

Explanation:

n = 2.419

Let the minimum angle of incidence is i.

The value of minimum angle of incidence so that all the light reflects back into the diamond is the critical angle for air diamond interface.

The relation for the critical angle and the refractive index of diamond is given by

Sin ic = 1 / n

where, ic is the critical angle for air diamond interface and n be the refractive index for diamond.

Sin ic = 1 / 2.419 = 0.4133

ic = 24.4 degree

Thus, the value of minimum value of angle of incidence for which all the light reflects back into diamond is 24.4 degree.

Explanation:

A light ray is traveling in a diamond (n = 2.419). If the ray approaches the diamond-air interface, what is the minimum angle of incidence that will result ...

Final answer:

To calculate the power converted to thermal energy due to frictional effects, subtract the ideal mechanical power required to lift water to the higher reservoir (13.2435 kW) from the actual pump power (20 kW), yielding 6.7565 kW.

Explanation:

The student is asking to determine the mechanical power that is converted to thermal energy due to frictional effects when water is pumped from a lower reservoir to a higher reservoir. Given the shaft power of the pump is 20 kW, height difference is 45 m, and flow rate is 0.03 m³/s, we can first calculate the ideal mechanical power required to lift the water to that height.

The gravitational potential energy given to the water per second (which is the power) by pumping it to the height is calculated using the formula P = ρghQ, where ρ (rho) is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.81 m/s²), h is the height difference (45 m), and Q is the flow rate (0.03 m³/s). This results in P = 1000 kg/m³ * 9.81 m/s² * 45 m * 0.03 m³/s = 13243.5 W or 13.2435 kW.

The difference between the actual power supplied by the pump (20 kW) and the ideal power required (13.2435 kW) is the power lost to thermal energy due to friction. Therefore, the power converted to thermal energy is 20 kW - 13.2435 kW = 6.7565 kW.

Answer:

555

Explanation:

The scenario is shown in the image attached below. A Right Angled Triangle is formed. We have an angle of 6 degrees, the side adjacent to angle which measures 1 mile and we need to find the side opposite to the angle.

Since 1 mile = 5280 feet, the side adjacent to the angle has a measure of 5280 feet.

Tangent ratio relates the opposite and adjacent side by following formula:

[tex]tan(\theta)=\frac{Opposite}{Adjacent}[/tex]

Using the given values, we get:

[tex]tan(6)=\frac{x}{5280}\\\\ x=tan(6) \times 5280\\\\ x = 555[/tex]

Thus, rounded to nearest foot, the height of the building is 555 feet

To find the height of the building, we can use trigonometry and the given angle of elevation. The height of the building is approximately 555 feet.

To find the height of the building, we can use trigonometry and the given angle of elevation. Let x be the height of the building. From the given information, we have the opposite side (x) and the adjacent side (1 mile = 5280 feet). The tangent function is used to relate these sides: tan(6°) = x/5280 feet. Solving for x, we get x = 5280 feet × tan(6°).

Using a calculator, we find that tan(6°) is approximately 0.1051. Multiplying this by 5280 feet gives us the height of the building in feet: 0.1051 × 5280 = 554.928 feet.

Rounding to the nearest foot, the height of the building is approximately 555 feet.

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Answer:

Yes we are right as the force on wire is approx 12500 Lb

Explanation:

Magnetic force on a current carrying bar is given by the equation

[tex]\vec F = i(\vec L \times \vec B)[/tex]

here we know that

L = 1.3 m

B = 12 T

[tex]\theta = 60^0[/tex]

i = 4.1 kA

now from above formula we have

[tex]F = iLBsin60[/tex]

[tex]F = (4.1\times 10^3)(1.3 )(12)sin60[/tex]

[tex]F = 55391 N[/tex]

So this is equivalent to 12500 Lb force

 Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.

"The magnetic force exerted on the conducting bar can be calculated using the formula for the force on a current-carrying conductor in a magnetic field, which is given by [tex]\( F = ILB \sin(\theta) \),[/tex] where [tex]\( I \)[/tex] is the current, \( [tex]L \)[/tex] is the length of the conductor, [tex]\( B \)[/tex]is the magnetic field strength, and [tex]\( \theta \)[/tex] is the angle between the current direction and the magnetic field.

 Given:

[tex]- \( I = 4.1 \times 10^3 \) A (since 4.1 kA = 4.1 \(\times\) 10^3 A) - \( L = 1.3 \) m\\ - \( B = 12 \) T\\ \( \theta = 60^\circ \)\[/tex]

Converting the angle from degrees to radians (since the sine function in physics formulas typically uses radians), we have[tex]\( \theta = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \) radians.[/tex]

 Now, we can calculate the force:

[tex]\( F = ILB \sin(\theta) \)[/tex]

[tex]\( F = (4.1 \times 10^3 \text{ A}) \times (1.3 \text{ m}) \times (12 \text{ T}) \times \sin\left(\frac{\pi}{3}\right) \)[/tex]

 Since[tex]\( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex], we can substitute this value into the equation:

[tex]\( F = (4.1 \times 10^3) \times (1.3) \times (12) \times \frac{\sqrt{3}}{2} \)[/tex]

 Calculating the force in newtons:

[tex]\( F = 4.1 \times 10^3 \times 1.3 \times 12 \times \frac{\sqrt{3}}{2} \)[/tex]

[tex]\( F \approx 4.1 \times 1.3 \times 12 \times 0.866 \times 10^3 \)[/tex]

[tex]\( F \approx 5107.2 \times 0.866 \)[/tex]

[tex]\( F \approx 4424.4 \) N[/tex]

 To convert the force from newtons to pounds, we use the conversion factor [tex]\( 1 \text{ N} \approx 0.22481 \text{ lb} \):[/tex]

[tex]\( F \approx 4424.4 \text{ N} \times 0.22481 \frac{\text{lb}}{\text{N}} \)[/tex]

[tex]\( F \approx 994.5 \text{ lb} \)[/tex]

 Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.

 In conclusion, the claim that the magnetic force will exceed 10,000 pounds is incorrect. The actual force is approximately 994.5 pounds, which is less than 10,000 pounds. The conducting bar should still be clamped in place to prevent movement due to the magnetic force, but the initial claim overestimated the force."

Answer:57.7

Explanation:

Force on a moving charge in  a magnetic field is given by

[tex]F=q\times V\times Bsin\theta [/tex]

Where F=Force experienced by charge

q=charge of particle

V=velocity of particle

B=magnetic field

F=qVBsin(25)-------1

[tex]2F=qVBsin(\theta )------2[/tex]

Divide (1)&(2)

[tex]\frac{1}{2}[/tex]=[tex]\frac{sin(25)}{sin(\theta)}[/tex]

[tex]Sin(\theta )=2\times sin(25)[/tex]

[tex]\theta =57.7 ^{\circ}[/tex]

Given the constraints of the question and the formula for magnetic force on a moving charge, the situation described isn't possible.

The magnetic force experienced by a moving charged particle is given by the formula F = qvBsin(θ), where q is the charge, v is the speed, B is the magnetic field strength, and θ is the angle between the velocity of the charge and the magnetic field. When the force is F at 25°, to experience a force of 2F, the sin of the new angle must be double that of sin(25°) because all other factors (charge, speed, magnetic field strength) are constant.

In the case given, sin(θ) would need to be 2sin(25°), but since the maximimum value sin(θ) can have is 1 (at 90°), this situation isn't possible within the constraints of the question (angle less than 90°). Hence, the student might have misunderstood the problem or there might be an error the question itself. However, if it is assumed that the question intends to find when the component of the force perpendicular to the field is double, the answer would be when θ is 90 degrees because the sin(90°) = 1. This is because when a charged particle moves perpendicular to a magnetic field, the sin(θ) factor in the force equation is maximized.

To find the magnitude of the induced current in the coil at 0.10 s, we need to calculate the magnitude of the magnetic flux through the coil and use Faraday's law of electromagnetic induction.

To find the magnitude of the induced current in the coil at 0.10 s, we need to first calculate the magnitude of the magnetic flux through the coil. The magnetic flux is given by Φ = B * A, where B is the magnetic field and A is the area of the coil. The area of the coil is calculated as A = π * r^2, where r is the radius of the coil.

Next, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. Mathematically, this can be written as emf = -dΦ/dt. Since the coil has resistance, we can use Ohm's Law to find the magnitude of the induced current, which is given by I = emf/R. Plugging in the values and calculating, we can find the magnitude of the induced current at 0.10 s.

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To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction. The magnetic field B is given as B = 50sin(10πt) mT. At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction, which states that the induced emf (voltage) is equal to the negative rate of change of magnetic flux through the coil.

The magnetic field B is given as B = 50sin(10πt) mT, where t is measured in seconds.

The magnetic flux through the coil is given by Φ = B.A, where A is the area of the coil.

Substituting the given values, we have A = πr^2 = π(0.04^2) = 0.005 cm^2.

At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.

We can now calculate the magnetic flux through the coil using Φ = B.A.

Finally, we can use Faraday's Law to find the induced emf and divide it by the resistance of the coil to find the magnitude of the induced current.

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Answer:

7212.3 N

Explanation:

F = 7.38 x 10^4 N, v = 36.2 m/s

Let b be the strength of magnetic field and charge on the particle is q.

F = q v B Sin theta

Here theta = 90 degree

7.38 x 10^4 = q x 36.2 x B x Sin 90

q B = 2038.7 .....(1)

Now, theta = 17 degree, v = 12.1 m/s

F = q v B Sin theta

F = 2038.7 x 12.1 x Sin 17     ( q v = 2038.7 from equation (1)

F = 7212.3 N

Answer:

There  will be no voltage across  secondary coil because DC voltage source is used in primary coil so there is no electromotive force induced in secondary coil.

Explanation:

In this question we have given

AA battery is used to supplies a constant voltage of 1.5 volts to primary coil of  transformer. In this case, voltage source is DC source which is providing constant voltage and for a transformer to work it is necessary to use an AC source.

Therefore, no EMF will induce in the secondary coil

and we know that in a transformer,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1} }=\frac{N_{2} }{N_{1} }[/tex]...........(1)

hence from above equation it is clear that,

[tex]\frac{E_{2} }{E_{1}} =\frac{V_{2} }{V_{1}}[/tex]............(2)

Here ,

[tex]E_{2}=0[/tex]

Put [tex]E_{2}=0[/tex] in equation (2)

We got

[tex]V_{2}=0[/tex]

There  will be no voltage across  secondary coil because there is no electromotive force induced in secondary coil.