what is the magnitude of its temperature change |δt|= 18.6 k in degrees celsius? view available hint(s)

Answer:

speed = 65.7 km/h

Explanation:

speed = distance over time, so you'd do 460 km/7 hrs to get your speed in km/h

The strength of the magnetic field in the center of the system, which is equally spaced from the wires. . B = μ0I/(2πr) for the magnetic field's strength.

And this is where the second important characteristic comes into play. The equivalent surface current density I = dI/ds =B = μ0I/(2πr), which is proportional to the magnetic field Br of the permanent magnet material, may be used to compute the magnetic field of permanent magnets. In this case, the permeability of open space is μ = 4 107 A/Tm.

The magnetic field lines encircle a long, straight wire when it is carrying a current I. You may find the magnetic field direction by circling the wire with your fingers while pointing your right thumb in the direction of the current.

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The speed of the Earth around the Sun is approximately 30 kilometers per second (km/s), or about 18.5 miles per second (mi/s).

The Earth orbits the Sun in an elliptical path with an average distance of about 149.6 million kilometers (93 million miles). The speed of the Earth in its orbit is not constant throughout the year but varies slightly due to the elliptical shape of the orbit and the gravitational influence of other celestial bodies. On average, the Earth's orbital speed is approximately 30 km/s or 18.5 mi/s. This orbital speed is a result of the balance between the gravitational attraction of the Sun and the centrifugal force acting on the Earth. The gravitational force pulls the Earth inward, while the centrifugal force pushes it outward. The Earth's orbital speed allows it to maintain a stable orbit around the Sun, where the gravitational force and the centrifugal force are in equilibrium.

It's worth noting that the speed of the Earth around the Sun is just one component of its overall motion in the universe. The Earth is also part of the solar system's rotation around the center of the Milky Way galaxy and the galaxy's motion within the expanding universe.

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Given the graph of period vs amplitude.

It can be seen from the graph that the plot of the period versus amplitude is almost a horizontal line. That is, as the amplitude increase, the period of the pendulum remains almost unchanged.

This is also consistent with the fact that the period of the simple pendulum indeed does not depend upon its amplitude.

Thus, according to the graph, the period does not depend on the amplitude.

The average thrust of the rocket engine to burnout is 4308.8 kN, the burnout time of the rocket is 140 seconds, and the speed at burnout is 8.91 km/s.

The force generated by the engine of the rocket is thrust. For the movement of the rocket, thrust is required. The rocket's mass and the acceleration it achieves as a result of the thrust are directly proportional.The initial weight of the single stage rocket is 500 kN, of which 420 kN is fuel, the weight of payload weight is 40 kN and the structural mass of the booster is 40 kN.Thus, the weight of the rocket is the sum of the weight of the fuel, the weight of the payload, and the weight of the structural mass of the rocket, which is 420 + 40 + 40 = 500 kN. Since the rocket starts from rest, the initial velocity of the rocket is zero and the distance covered by the rocket is also zero. Thus, the initial acceleration of the rocket is zero. When the rocket engine is ignited, it exerts a force called thrust, which is equal to the rate at which the rocket engine expels mass multiplied by the exhaust velocity. The mass of fuel ejected per second is the fuel burn rate, which is 3 kN/s. The exhaust speed is 3 km/s, so the thrust exerted by the rocket engine is given by the product of fuel burn rate and exhaust speed. Thus, the average thrust of the rocket engine to burnout can be determined by the formula: T = * + / * , where T is the average thrust of the rocket engine to burnout, is the mass of the rocket at any time t, is the acceleration due to gravity, is the exhaust velocity, and is the time taken to burn the entire fuel. = 500 – 3 (mass of the rocket after time t), = 9.6 m/s², and = 3 km/s or 3000 m/sThus, T = (500 – 3) * 9.6 + 3 * 10³ = 4800 – 28.8 + 3000The burnout occurs when the fuel runs out. = 500 – 3 = 0t = 500/3 seconds = 166.67 seconds Therefore, the average thrust of the rocket engine to burnout is T = 4800 – 28.8(166.67) + 3000 = 4308.8 kN. The burn time of the rocket is the time it takes to burn all of the fuel and reach burnout. The rocket's burnout time can be determined using the fuel burn rate, which is 3 kN/s and the mass of fuel in the rocket, which is 420 kN. Since the mass of fuel ejected per second is constant, the burn time of the rocket is obtained by dividing the mass of fuel by the fuel burn rate. = 420 kN, / = 3 kN/st = // = 140 seconds Thus, the burnout time of the rocket is 140 seconds

The velocity of the rocket at burnout is obtained by integrating the thrust of the rocket engine with respect to time. = 500 – 3 = 80 kN, = 9.6 m/s², and = 3 km/s or 3000 m/sT = * + / * = 768 + 240000/The velocity of the rocket at burnout is the integral of the thrust from zero to the time of burnout, divided by the mass of the rocket at burnout. = 1/80 * ∫₀¹⁴₀ (768 + 240000/) = 8.91 km/s Thus, the speed at burnout is 8.91 km/s.

The average thrust of the rocket engine to burnout is 4308.8 kN, the burnout time of the rocket is 140 seconds, and the speed at burnout is 8.91 km/s.

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Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

\(x=4t^3+3t^2-5t+2\)

Where,

x is in meters and t is in sec

We know that,

Velocity,

\(v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5\)

(a) i. t = 2 s

\(v=12(2)^2+6(2)-5=55\ m/s\)

At t = 4 s

\(v=12(4)^2+6(4)-5=211\ m/s\)

(b) Acceleration,

\(a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6\)

Pu t = 3 s in the above equation

So,

\(a=24(3)+6\\\\a=78\ m/s^2\)

Hence, this is the required solution.

Hence, a 200.7 joule change in the box's potential energy has occurred (J).

Potential energy has a simple definition: It is a type of energy that possesses the ability to perform work but is not now performing work or exerting any force on any other objects. In other words, potential energy concerns object positions rather than object motion.

The following equation provides the potential energy (PE) gained by lifting an object against gravity:

PE = m * g * h

where m is the object's mass, g is its gravitational acceleration, and h is the height to which it is raised.

Substituting the given values, we get:

PE = 11 kg * 9.81 m/s² * 1.7 m

PE = 200.7 J

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The statement "Lt= L1 + L2 + L3 is the formula for finding the total inductance of a circuit when there is more than one inductor connected in series" is True.

The formula Lt = L1 + L2 + L3 is the correct formula for finding the total inductance of a circuit when there is more than one inductor connected in series.

When inductors are connected in series, the total inductance of the circuit is the sum of the individual inductances of the inductors. This is because the magnetic field created by each inductor adds together and the effective inductance is the sum of the individual inductances.

This is based on the principle of additivity of inductances in series. The total inductance of a series circuit is equal to the sum of the individual inductances. The current flowing through each inductor is the same and the voltage across each inductor is proportional to its inductance.

It is important to note that this formula is only true for inductors connected in series. When inductors are connected in parallel, the total inductance is given by the formula:

1/Lt = 1/L1 + 1/L2 + 1/L3 + ...

Where L1, L2 and L3 are the inductances of the inductors.

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Answer:

The magnitude of the student's displacement is less than the distance she walked.

Walking:  11 Blocks

Displacement:  8.54 Blocks

Explanation:

See the attached diagram.  The unit of length is blocks.  We can add the actual blocks walked as shown.  She walked a total of 11 blocks.

Her displacement is the distance measured directly from where she started (line A).  Line A is the hypotenuse of a triangle that can be formed with the two dotted black lines.  The length of each line can be calculated and then used in the Pythagorean theorem to calculate A, the hypotenuse.

That result is 8.54 Blocks, a shorter distance, once she earns her wings.

If a student on her way to school walks four blockers east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than the total distance walked by her.

An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard.

As given in the problem if a  student on her way to school walks four blockers east, three blocks north, and another four blocks east,

The total distance walked by the student = 11 blocks

Displacement of the student from home = √(8² + 3²)

                                                                   = 8.5 blocks

The total displacement by the student would be less than the

Thus, the magnitude of her displacement from home is less than the distance.

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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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Answer:

The answer is Radiation.

Explaination:

A and B are wrong because you can't have convection or convection currents without air, and all the air has been removed from the box.

D is wrong because there is no liquid in the box.

C is correct because heat can be transmitted even through a vacuum by radiation (e.g., the Sun). You can quickly feel the heat radiated by a piece of white-hot iron.

Whether or not each type of glassware can be heated depends on the specific type of glassware and the method of heating.

The method of heating refers to the various ways in which heat energy can be transferred to an object or substance to raise its temperature. Some common methods of heating include:

Whether or not each type of glassware can be heated depends on the specific type of glassware and the method of heating. Some types of glassware, such as borosilicate glass, are designed to withstand high temperatures and are commonly used for heating applications, while other types of glassware may be more susceptible to cracking or breaking when exposed to heat.

Glassware that is designed for heating applications will be labeled as such and will typically have a higher temperature tolerance than glassware that is not intended for heating. Some common types of glassware that are designed for heating include Pyrex, Kimax and Vycor glass.

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For most everyday objects, scientists use a balance to obtain an object's mass. A balance compares an object with a known mass to the object in question. One example of a balance is the triple beam balance.

It will take 2 seconds to reach 30 m/s north.

Below are the parameter given from the question :

Initial velocity U = 20 m/s

Final velocity V = 30 m/s

Acceleration a = 5 m/s

Time t = ?

The equation of linear motion to use is:

V = U + at

Substitute all the parameter to the equation above.

30 = 20 + 5t

Collect the like term

5t = 30 - 20

5t = 10

t = 10/5

t = 2 s

Therefore, It will take 2 seconds to reach 30 m/s north.

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Answer:

101.937 kg

Explanation:

The force needed to get the box moving must just cancel the static friction force:

F = µsmg  =  0.35•m•9.81 = 350 --->   m = 350 / (0.35•9.81) = 101.937 kg

Again, with units shown, and using 1 N = 1 kg•m/s2:

0.35•m•9.81(m/s2) = 350 N

solving for m:

m = 350 N / (0.35•9.81 m/s2)  =  (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg

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