a round flat metal coil has 180 turns and negligible resistance. it is connected in a series circuit with a resistor, with nothing else in the circuit. you measure that a current flows through the resistor when a magnetic field through the coil, perpendicular to its area, is changing at what is the radius of the coil?

For people who don't speak Spanish.

translation:  

Identify the type of variable for each given situation (qualitative: nominal or ordinal) or (quantitative: continuous or discrete)

sorry though I cant answer your question.

The critical angle of the transparent material is 35.3 degrees.

The critical angle is the angle of incidence at which the angle of refraction is 90 degrees. In this case, the angle of refraction is 32 degrees. Therefore, the critical angle is calculated as follows:

sin(critical angle) = sin(90 degrees) / sin(angle of refraction)

sin(critical angle) = 1 / sin(32 degrees)

sin(critical angle) = 0.574

critical angle = arcsin(0.574)

critical angle = 35.3 degrees

Therefore, the critical angle of the transparent material is 35.3 degrees.

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The correct answer is opposite to the displacement.

The spring force is the force which is exerted by the spring when it is attached with an object.

When we attach a spring to an object and the object is displaced from equilibrium, then there will be a force acting on the spring which will retract the spring to come back to equilibrium.

This is spring force.

According to the Newton's third law, there is always an equal and opposite force. Here, if we are applying a force to displace the object. This force is applied on the object so it is towards the object. Then there is an equal and opposite force which is applied on the spring which tends to pull the object back to its equilibrium position. This force which acts opposite to the displacement of the object to bring spring back to equilibrium is the spring force.

Therefore, we can conclude that, in general, the spring force always points opposite to the displacement from equilibrium.

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Answer:

7 m

Explanation:

They began walking 6 meters to the right, so +6.

Then 3 meters left so -3,

And then 4 meters to the right again so +4.

4+6-3=7 meters

The height of the ceiling at the center of the whispering gallery is approximately 43.3 feet.

In an ellipse, the sum of the distances from any point on the ellipse to its two foci is constant. In this case, the two foci are located 25 feet from the center of the hall.

Given that the hall is 100 feet in length, the distance from one end to the center is 50 feet. We can consider this as the semi-major axis (a) of the ellipse.

The sum of the distances from any point on the ellipse to its two foci is equal to 2a. Thus, the sum of the distances from the ceiling at the center of the hall to the two foci is also 2a.

Since the foci are located 25 feet from the center, the sum of the distances is 2a = 50 feet.

To find the height of the ceiling at the center, we need to determine the semi-minor axis (b) of the ellipse. The semi-minor axis can be calculated using the formula:

b = √(a² - c²)

where c represents the distance from the center to each focus. In this case, c = 25 feet.

Substituting the values into the formula:

b = √(50² - 25²)

b = √(2500 - 625)

b = √(1875)

b = 43.3 feet

Therefore, the height of the ceiling at the center of the whispering gallery is approximately 43.3 feet.

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Answer:

kilograms and m²/s²

Explanation:

The SI units combined to describe energy is kilograms and m²/s² ;

  Energy can be derived from;

Potential energy  = mgh

   Units               = kg x m/s² x m

                          = kg m²/s²

Even if we use kinetic energy, we still get kg m²/s²

Answer:

The speed of the airplane after it has travelled 800 meters is approximately 85.632 meters per second.

Explanation:

From statement, we obtain the equation of equilibrium associated with the airplane:

\(\Sigma F = T - R = m\cdot a\) (1)

Where:

\(T\) - Thrust force from engines, measured in newtons.

\(R\) - Force of air resistance, measured in newtons.

\(m\) - Mass of the airplane, measured in kilograms.

\(a\) - Net acceleration of the airplane, measured in meters per square second.

If we know that \(T = 7.5\times 10^{4}\,N\), \(R = 4.0\times 10^{4}\,N\) and \(m = 1.5\times 10^{4}\,kg\), then the net acceleration of the airplane is:

\(a = \frac{T-R}{m}\)

\(a = \frac{7.5\times 10^{4}\,N-4.0\times 10^{4}\,N}{1.5\times 10^{4}\,kg}\)

\(a = 2.333\,\frac{m}{s^{2}}\)

Lastly, the final speed of the airplane is calculated by the following kinematic equation:

\(v_{f} = \sqrt{v_{o}^{2}+2\cdot a \cdot \Delta s}\) (2)

Where:

\(\Delta s\) - Travelled distance, measured in meters.

\(v_{o}\), \(v_{f}\) - Initial and final speeds of the aircraft, measured in meters per second.

If we know that \(v_{o} = 60\,\frac{m}{s}\), \(a = 2.333\,\frac{m}{s^{2}}\) and \(\Delta s = 800\,m\), then the final speed of the aircraft is:

\(v_{f} = \sqrt{\left(60\,\frac{m}{s} \right)^{2}+2\cdot \left(2.333\,\frac{m}{s^{2}} \right)\cdot (800\,m)}\)

\(v_{f} \approx 85.632\,\frac{m}{s}\)

The speed of the airplane after it has travelled 800 meters is approximately 85.632 meters per second.

The electric field strength at a distance of 1.20 m on the x-axis is -1.5 × 10⁴ N/C.

To find the electric field strength at a distance 1.20 m on the x-axis, we can use Coulomb's law:

\($$F=k\frac{q_1q_2}{r^2}$$\)

where F is the force between two charges, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is the Coulomb constant.For a single point charge q located at the origin of the x-axis, the electric field E at a distance r is given by:

\($$E=\frac{kq}{r^2}$$\)  where k is the Coulomb constant.

So, let's calculate the electric field due to each charge separately and then add them up:

For the +2.0 C charge at x = 0, the electric field at a distance of 1.20 m is:\($$E_1=\frac{kq_1}{r^2}=\frac{(9\times10^9)(2.0)}{(1.2)^2}N/C$$\)

For the -3.0 C charge at x = 0.40 m, the electric field at a distance of 1.20 m is:

\($$E_2=\frac{kq_2}{r^2}\)

\(=\frac{(9\times10^9)(-3.0)}{(1.20-0.40)^2}N/C$$\)

The negative sign indicates that the direction of the electric field is opposite to that of the positive charge at x = 0.

To find the net electric field, we add the two electric fields\(:$$E_{net}=E_1+E_2$$\)

Substituting the values of E1 and E2:

\($$E_{net}=\frac{(9\times10^9)(2.0)}{(1.2)^2}-\frac{(9\times10^9)(3.0)}{(0.8)^2}N/C$$E\)

net comes out to be -1.5×10⁴ N/C.

Therefore, the electric field strength at a distance of 1.20 m on the x-axis is -1.5 × 10⁴ N/C.

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The orbital period of the satellite circling the Earth at an altitude of 3500 km is 163 minutes

The orbital period for the satellite circling the Earth at an altitude of 3500 km can be obtained as follow:

T² = (4π² / GM) × a³

T² = [(4 × 3.14²) / (6.67×10¯¹¹ × 5.987×10²⁴)] × 9900000³

Take the square root of both sides

T = √[((4 × 3.14²) / (6.67×10¯¹¹ × 5.987×10²⁴)) × 9900000³]

T = 9789.15 s

Divide by 60 to express in minutes

T = 9789.15 / 60

T = 163 minutes

Thus, we can conclude that the orbital period of the satellite is 163 minutes

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so the resulting force of all the 3 forces is 11.7 N.

As shown in the figure below ther is 3 forces \(F_1 ,F_2 ,F_3\) are the forces so that the diagram is along the directions given in the question

so \(F_1\) = 3 j^ ( where j^ is unit vector along y axis)

\(F_2\) = 6 cos 45  i^ + 6 sin 45 j^ ( where i^ is unit vector along x axis)

\(F_3\)  = - 5 sin 30 i^ + 5 cos 30 j^

so the resultant force = \(F_1+ F_2 +F_3\)

=> 3 j^ + 4.24 i^ + 4.24 j^ + 4.33 j^ - 2.5 i^

=> 1.74 i^ + 11.57 j^

so resultant force = \(\sqrt{1.74^2 + 11.57^2}\)

=> \(\sqrt{136.89}\)

=> 11.7 N

so the resulting force of all the 3 forces is 11.7 N

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To test the circuit breaker's tripping behavior, you can perform an overload test.

The following are the procedures for testing the circuit breaker:

Inspection:

Examine the circuit breaker for any visual harm or defects.

Ensure that all screws and terminals are secure and free of debris.

Continuity:

Use a multimeter to check the circuit breaker's continuity.

Ensure that there is continuity between the input and output wires of the circuit breaker and that the multimeter's resistance reading is zero when the breaker is closed.

Load:

Examine the circuit breaker's load-carrying capacity.

To do so, use a power source that is within the circuit breaker's current range and attach the circuit breaker to it.

As a result, the power source's current would flow through the circuit breaker.

Observe the tripping behavior:

After the power source has been connected, increase the current gradually to see if the circuit breaker trips at the appropriate value.

To do this, use a variable resistor or a load bank device to regulate the current flowing through the circuit breaker.

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Answer:

The sun is located in the Milky Way Galaxy.

Unfortunately, without the value of theta, we cannot find the exact tension in the left chain. However, if you provide the angle between the chain and the vertical line, we can calculate the tension in the left chain.

To find the tension in the left chain of the swing, we need to consider the forces acting on the person sitting on the swing. Since the person weighs 120 N, the total vertical force should balance the person's weight. The tensions in the chains have both vertical and horizontal components. Let's focus on the vertical components.

Let T_left and T_right represent the tensions in the left and right chains, respectively. We know T_right = 200 N. As both chains are at equal angles, their vertical components can be represented as T_left * cos(theta) and T_right * cos(theta), where theta is the angle between the chain and the vertical line.

Now, we can set up an equation to represent the balance of the vertical forces:

T_left * cos(theta) + T_right * cos(theta) = 120 N

Since T_right = 200 N, we can substitute:

T_left * cos(theta) + 200 * cos(theta) = 120 N

Now, to find T_left, we need to factor out cos(theta):

cos(theta) * (T_left + 200) = 120 N

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One type of supplement that can improve exercise performance and capacity is Creatine.

What is Creatine? Creatine is a nitrogen-containing compound that's naturally present in the muscles. The compound is essential for the production of ATP, which is the body's main source of energy for muscular contractions.

Creatine supplements are known for their ability to enhance athletic performance and boost muscle mass and strength. Studies show that consuming creatine monohydrate supplements can improve exercise performance and increase muscle strength and mass in people engaging in high-intensity exercise regimes and resistance training. Creatine supplements are commonly used by athletes, bodybuilders, and weightlifters.

They are also considered to be one of the most well-researched and safest supplements that can improve exercise performance and capacity.

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Answer:

The last option.

Explanation:

Since you are going down, the gravitational potential energy would go down too. Thus, the gravitational potential energy decreases.

Since the gravitational potential energy is converted to kinetic energy when you move down, there is an increase in kinetic energy.

Answer:

The fourth option

Explanation:

Answer:

C

Explanation:

Pascals are the units for measuring pressure

U₂ = 0 V. The lamp (L₂) is short circuited, meaning that there is no resistance between its terminals. As such, the voltage across its terminals must be zero.

Resistance is a concept in physics that measures the opposition to the flow of electric current, heat, light, sound, or any other form of energy or matter. It is the measure of how much an object, material, or system resists the flow of electric current, heat, light, sound, or any other form of energy or matter.

a) U₂ = 0 V. The lamp (L₂) is short circuited, meaning that there is no resistance between its terminals. As such, the voltage across its terminals must be zero.

b) U₁ = 12 V. Since the generator is providing a constant voltage of 24 V across its terminals, the voltage across the terminals of (L₁) will be half of this, i.e. 12 V. As the lamps are identical, the current I' through the circuit will also be 0.6 A.

c) (L₁) may burn out due to the large amount of current it is carrying. If the current is too high, it may exceed the rated capacity of the lamp and lead to its failure.

d) (L₁) does not burn out because the circuit includes a fuse. The fuse is designed to break the circuit in the event of an excessive current, thus protecting the lamp from burning out. When the fuse blows, the circuit is broken and the lamp will be off.

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Answer:

30J

Explanation:

Given parameters:

Mass of box  = 1.5kg

Height  = 2m

Acceleration due to gravity  = 10N/kg

Unknown:

Change in gravitational potential energy  = ?

Solution:

 The change in gravitational potential energy is given as:

      ΔGPE  = mgh  

m is the mass

g is the acceleration due to gravity

h is the height

   ΔGPE  = 1.5 x 10 x 2  = 30J

The force by Earth on the Moon would decrease to one-ninth (1/9) of its original value if the distance between them were tripled.

According to Newton's law of universal gravitation, the force between two objects is inversely proportional to the square of the distance between them. If the distance between the Earth and the Moon were to triple, meaning the Moon is three times further from Earth than its current position, the force by Earth on the Moon would decrease.

The relationship between the force (F) and the distance (r) can be expressed as:

F ∝ 1/r^2

If the distance triples, the new distance (r') would be three times the original distance (r). So, we can substitute r' = 3r into the equation:

F' ∝ 1/(3r)^2

F' ∝ 1/9r^2

F' = 1/9 * F

Therefore, the force by Earth on the Moon would decrease to one-ninth (1/9) of its original value if the distance between them were tripled.

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The two pieces of paper land on the moon's surface at the same time because they are both subject to the same gravitational force.

The force of gravity is determined by the mass of the object and the distance between it and the center of the celestial body. On the moon, the force of gravity is weaker than it is on Earth, but it is still strong enough to cause objects to fall toward the surface.

The shape of the paper, whether crumpled or folded into an airplane, has no effect on the force of gravity acting on it. In the absence of air resistance, all objects, regardless of their mass, shape, or size, will fall at the same rate in a vacuum.

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823 ft high is the stone 3 seconds later.

What is motion?

The change in the position of the object with respect to time is called motion in physics.

We know that

\(y=y_{0} + V_{0}-\frac{1}{2} gt^{2}\)

where,

\(y\) is the height of the stone at 3s (the value we want to find)

\(y_{0}\) is the initial height of the stone

\(V_{0}\) is the initial velocity of the stone

\(t\) is the time  at which we need to find the height

\(g\) is the acceleration due to gravity

Therefore,

\(y = 925+(14)(3)-\frac{1}{2} (32)(3)^{2}\)

= 823 ft

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The ball should be tossed with speed 10 m/s in order for it to take 2 second to return to the level from which it was tossed.

The Time of flight of projectile motion is defined as the total time taken during ascent and fall to the ground.

Here, time of flight is given 2 seconds. So, Time of ascent is 1 second. Time of ascent is time taken to reach the highest height.

We have to apply the formula

v= u +at

At the maximum height, v =0 m/s.

0 = u - 10×1

So, u = 10 m/s

So, our answer depends on the value of g. If we take the value of g as 9.8, then speed required will be 9.8 m/s

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The correct conductor ampacity for a 10-horsepower, 230-volt, single-phase motor is option c. 62.5 A.

The ampacity of a conductor is a measure of the maximum amount of electrical current that can flow through it safely. The ampacity required for a motor is based on the size of the motor and the voltage of the electrical system. There are standard methods for determining the minimum ampacity for a given motor and voltage, such as the National Electric Code (NEC) and the American National Standards Institute (ANSI). The NEC and ANSI both state that the minimum ampacity for a 10-horsepower, 230-volt, single-phase motor is 62.5 amperes. It's important to note that using a conductor with ampacity lower than the recommended one may cause the motor to overheat and fail, leading to damage or even fire hazard.

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Answer:

d. Both kinetic and sliding friction

Explanation:

Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.

Answer:

a) T=549.36N Upwards

b) T=448.56N Upwards

c) T=650.16N Upwards

Explanation:

The very first thing we can do to solve this problem is to draw a free body diagram we can use to analyze the situation (see attached picture).

On the diagram we can see there are only two forces acting on the object: the tension of the rope and the weight of the object itself.

a)

Since the object is moving at a constant speed, this means that its acceleration will be zero. So we can do a sum of forces like this:

\(\sum F=0\)

T-W=0

T=W

T=mg

\(T=(56kg)(9.81m/s^{2})\)

T=549.36N upwards

b)

For part b, since the object is accelerating downwards, we wil say that it's acceleration is negative, so \(a=-1.8m/s^{2}\)

so we can do a sum of forces again:

\(\sum F=ma\)

so

T-W=ma

T=ma +W

T=ma+mg

T=m(a+g)

and now we substitute:

\(T=(56kg)(-1.8 m/s^{2}+9.81 m/s^{2})\)

which yields:

T=448.56N upwards (in this particular case, the tension always goes upwards)

c)

Since the object is moving upwards, we can say that its acceleration will be positive, so \(a =1.8m/s^{2}\)

we can take the solved equation we got on the previous part of the problem, so we get:

T=m(a+g)

\(T =(56kg)(1.8 m/s^{2}+9.81 m/s^{2})\)

which yields:

T=650.16N upwards

The tension force in the rope when the object moves upward at a constant velocity is 548.8 N.

The tension force in the rope when the object accelerates downward at a constant acceleration is 448 N.

The tension force in the rope when the object accelerates upward at a constant acceleration is 649.6 N.

The given parameters:

The tension force in the rope when the object moves upward at a constant velocity;

T = ma + mg

at constant velocity, a = 0

T = mg

T = 56(9.8)

T = 548.8 N

The tension force in the rope when the object accelerates downward at a constant acceleration of 1.8 m/s².

T = mg - ma

T = m(g - a)

T = 56(9.8 - 1.8)

T = 448 N

The tension force in the rope when the object accelerates upward at a constant acceleration of 1.8 m/s².

T = mg + ma

T = m(g + a)

T = 56(9.8 + 1.8)

T = 649.6 N

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Based on the experiment conducted, we can infer the following true statements:

A wave can be defined as a disturbance in a medium that progressively transfers (transports) energy from a source to another location, especially without the transportation of mass (matter).

The type of wave that was created is a transverse wave because the medium went up and down.

The estimated wavelength of this transverse wave was equal to 20 meters.

The estimated frequency of this transverse wave was equal to 60 meters.

For the velocity, we have:

Velocity = wavelength × frequency

Velocity = 20 × 60

Velocity = 1200 m/s

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The correct option is (e) 1,020 N/C.

The maximum value of the electric field at the upper atmosphere of the earth can be calculated using the formula:

E = c*sqrt(I/(2*ε0))

Where c is the speed of light, I is the intensity of sunlight, and ε0 is the permittivity of free space.

Substituting the given values, we get:

E = 3 x 10^8 m/s * sqrt(1,380 W/m^2 / (2 * 8.85 x 10^-12 F/m))

E ≈ 1,020 N/C

Therefore, the maximum value of the electric field at the upper atmosphere of the earth is approximately 1,020 N/C.

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If an air parcel cools adiabatically at the surface with a temperature of 17 °C, the temperature drops by about 10 °C for every 1000 meters of altitude. 500mb is typically about 5500 meters from the surface, so with dry adiabatic cooling the package temperature will be about 17°C - (10°C * 5.5) = -33°C.

On the other hand, if the air parcel is moist and adiabatically cooled, the temperature change will be slower because latent heat is released when the water vapor condenses. The adiabatic temperature drop rate when wet is typically about 6°C per 1000 meters.

However, it is generally cooler than the dry adiabatic decay rate, so for wet adiabatic cooling, the -15°C, -25°C, 0, and -35°C options are possible temperatures for a 500MB package. There is a possibility. . 

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