The maximum allowed power dissipation for a 27.3-ω resistor is stated to be 10.0 w. find the largest current that this resistor can take safely without burning out.

The location in space where Earth's and the Moon's gravitational attractions are equal is closer to the Moon due to its significantly lesser mass compared to Earth. By applying Newton's law of universal gravitation, it can be deduced that this point must be nearer the Moon, as Earth's stronger gravitational force diminishes with increasing distance.

The gravitational attraction between two objects depends on both their masses and the distance between them, according to Newton's law of gravitation. The force of gravity is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers of mass. Based on the data provided, the Earth's gravitational force is much stronger than that of the Moon due to its greater mass. However, as the distance from the Earth increases, its gravitational pull weakens.

Given that the mass of the Moon is about 1/81 of the Earth's mass and the distance from the Earth to the Moon is approximately 3.80×105 km, there exists a point where the gravitational forces exerted by Earth and the Moon on an object are equal, known as the Lagrange point L1. This point is closer to the Moon than to Earth because the Moon's weaker gravitational force requires a shorter distance to match the stronger gravitational force of the Earth. To find this exact point, one would use the formula from Newton's law of gravitation and set the forces equal to each other, solving for the distance from Earth at which this equilibrium occurs.

Considering the information provided about the distances and gravitational forces, we can infer that the space pod mentioned in the question, at the point where Earth's and the Moon's gravity cancel each other out, would indeed be closer to the Moon. This is a consequence of the vast difference in mass between the Earth and the Moon and the inverse-square law of gravity.

Final answer:

To find the internal resistance of a 12.0-V car battery with specific measurements, use Ohm's Law.

Explanation:

Internal resistance of a battery can be calculated using Ohm's Law, where internal resistance = (emf - terminal voltage) / current. In this case, with an EMF of 12.0 V, terminal voltage of 8.3 V, and current of 89 A, the internal resistance would be calculated as (12.0 V - 8.3 V) / 89 A.

This gives an internal resistance of approximately 0.0427 ohms.

The answer i got was 50 miles per hour

Answer: Option D

Explanation: The fertilizers consists of the phosphates and nitrates. These aids in the growth of the plant. Fertilizers used by the plants are sometimes carried away due to the rain.

The run off from the agricultural field contains phosphates that is transferred from the field to the nearby water bodies.

If the particle is under the Simple Harmonic Motion, the wave is considered to be sinusoidal wave.

Explanation:

A body is considered to be performing the Simple Harmonic Motion if it repeats its motion in a particular path with a specific time period. The two and fro motion of the pendulum about a fixed mean position is termed as the Simple Harmonic Motion.

The motion of the simple pendulum repeats itself after a particular time period and it continuously retraces its path during the motion. Therefore, it is termed as the simple harmonic motion.

The wave depicting the motion of a body performing the simple harmonic motion is shown in the sinusoidal manner. The motion of the body from its mean position to its extreme and then from extreme position to its mean position shows the positive cycle of the sinusoidal waveform.

The motion of the pendulum on the other side of the mean position of the pendulum represents the other half of the sinusoidal waveform.

Thus, If the particle is under the Simple Harmonic Motion, the wave is considered to be sinusoidal wave.

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Answer Details:

Grade: High School

Chapter: Simple Harmonic Motion

Subject: Physics

Keywords:

simple, harmonic, SHM, motion, pendulum, oscillates, mean position, extreme, repeats, time period, sinusoidal.

Answer:

the 3rd choice is the answer

Explanation:

The weight of an astronaut standing on the moon can be calculated using the equation weight = mass × acceleration due to gravity. By calculating the weight using the ratio of acceleration due to gravity on the moon compared to that on Earth, we find that an astronaut with a weight of 210 lb on Earth would weigh approximately 158.18 lb on the moon.

The weight of an astronaut standing on the moon can be calculated using Newton's second law of motion. The formula to calculate weight is weight = mass × acceleration due to gravity. Since the acceleration due to gravity on the moon is approximately 1/6 that at the surface of the earth, we can calculate the weight using this ratio.

First, we need to convert the weight on earth from pounds to kilograms. Since 1 kilogram is approximately 2.2 pounds, we divide the weight on earth (210 lb) by 2.2 to get the weight in kilograms.

Next, we multiply the weight in kilograms by the acceleration due to gravity on the moon (1/6 of 9.8 m/s^2) to find the weight of the astronaut on the moon.

This results in a weight of approximately 158.18 lb on the moon for an astronaut whose weight on earth is 210 lb.

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Answer: Opposite direction

Skeletal muscles arise in antagonistic pairs where the muscles contract to produce opposite movements at the same joint. Antagonistic muscles move a body part in one direction by contraction, the other moves that part in opposite direction.

In addition, when a muscle contracts to produced movement, its antagonist relaxes to allow movement to take place such as biceps muscles is a flexor muscle for elbow joint and triceps is the antagonist.

Answer: False

When the diaphragm contracts, the muscles will also contract and pull upward and increase the size of the thoracic cavity thus decreases air pressure inside during inspiration. After the diaphragm contracts, it goes to relaxation, the muscles will also relaxed. It gets looser and return to its original position higher up in the chest. This increase the pressure in the chest, which force the air in the lungs out through the nose.

Part A: The potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.

Part B: The distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.

Part C: The energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.

How do you calculate the potential?

Given that the diameter of the sphere is 2.00 m and the charge of 5.00 mc.

Part A

The potential on the surface of a charged sphere is given below.

[tex]V = k\dfrac {Q}{R}[/tex]

Where V is the potential on the surface, Q is the charge, R is the radius and k is the Coulomb's constant.

[tex]V = 8.99 \times 10^9\times \dfrac {5 \times 10^{-3}}{\dfrac {2}{2}}[/tex]

[tex]V = 4.5 \times 10^7 \;\rm Volts[/tex]

Hence the potential on the surface of the charged sphere is [tex]4.5 \times 10^7[/tex] volts.

Part B

The potential of 1 mv generated at a certain distance x from the centre of the sphere is given below.

[tex]V_x = k \dfrac {Q}{x}[/tex]

[tex]1 \times 10^{-3} = 8.99 \times 10^9 \times \dfrac{5 \times 10^{-3}} {x}[/tex]

[tex]x = 4.5 \times 10^{10} \;\rm m[/tex]

Hence the distance is [tex]4.5 \times 10^{10}[/tex] m from the center of the sphere where the potential is 1 mv.

Part C

Given that an oxygen atom with three missing electrons is released near the van de-Graaff generator. It means that the charge Q = +3e

The electric potential energy of a charged particle located at some point with voltage V is given below.

[tex]U = QV[/tex]

At the distance found in part b, which is [tex]x = 4.5 \times 10^{10} \;\rm m[/tex], the energy of the oxygen atom is,

[tex]U = +3e \times 1 \times 10^{-3}[/tex]

[tex]U = 3me \;\rm V[/tex]

Hence the energy of the oxygen atom found at the distance of [tex]x = 4.5 \times 10^{10} \;\rm m[/tex] is 3me V.

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The resistivity of two copper wires remains the same regardless of their lengths because resistivity is an intrinsic property of the material.

Comparing resistivities of two copper wires of different lengths but the same material, we need to understand that resistivity is an intrinsic property of the material and does not depend on the geometry of the wire. Therefore, regardless of whether one wire is twice as long as the other, the resistivity of both wires remains the same. This is a fundamental concept in the study of electrical resistance and materials.

The horizontal component of the velocity of a stone thrown upward at an angle remains constant during its flight, as there is no horizontal force to affect it. This is due to the principles of projectile motion in physics. Gravity only affects the vertical velocity, not the horizontal.

In physics, the motion of a projectile thrown at an angle can be separated into its vertical and horizontal components. Both of these components operate independently of each other. When a stone is thrown upward at an angle, the horizontal component of its velocity remains constant throughout the flight, under ideal conditions, because there's no horizontal force to cause acceleration or deceleration.

So, as the stone rises and falls, the horizontal velocity remains unchanged because gravity only affects the vertical component of the velocity. In this process known as projectile motion, gravity pulls the object downward, causing the vertical component of velocity to decrease as the stone rises, and increase as it falls, but the horizontal component is unaffected by gravity. The effect of air resistance is typically ignored in basic physics problems, but in reality, it would gradually decrease the stone's velocity in both directions.

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Answer: D

expends the least amount of energy for the greatest result

Explanation:

edge 2020

Answer:

C) TURBINE

Explanation:

As we know that wind powered system is used to generate electricity using wind energy.

Here wind powered system has different components

1) Blades: these are long plates connected to the turbine. when these plates encountered high speed wind they start rotating on its axle.

This will produce kinetic energy in the axle which is transferred to the turbine.

2) Turbine: It is used to convert mechanical energy of the wind into useful electrical energy.

This is continuously driven by the kinetic energy of wind

So here correct answer is

C) Turbine

ANSWER: Ian is demonstrating a typical example of Observational Learning. An observational learning is a learning pattern which happens by observing the behavior of others. People generally follows a role model such as parent, sibling, relative or teachers and imitate them while they are doing their work, particularly in childhood. There are four steps of observational learning which are Attention, Memory, Initiation and Motivation.

a dense substance that is hard and incompressible

for those on ~ a p e x ~ note that this question asks when the resistors are connected in SERIES

if you have the question for in PARALLEL the answer is:

> The current the battery equals to the sum of the currents in the resistors

wink wonk

The properties that Mg, Ca and K do not have in common is "same number of valence electrons."

In the periodic table, elements are classified into groups and periods. The groups of elements show the relationships between them in terms of reactivity, valence electrons and physical properties.

Elements in the same group have the same number of valence electrons. However, Ca, K and Mg are not in the same group so they can not have the same number of valence electrons. Ca and Mg has two valence electrons while K has only one valence electron.

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The velocity function can be found by differentiating the position function. The velocity after 1 second is -0.02 ft/s. The particle is at rest at t = 0 and t = 1.5 seconds.

(a) To find the velocity at time t, we can differentiate the position function. The derivative of f(t) = 0.01t^4 - 0.02t^3 is v(t) = 0.04t^3 - 0.06t^2. So, the velocity function is v(t) = 0.04t^3 - 0.06t^2 ft/s.

(b) To find the velocity after 1 second, we substitute t = 1 into the velocity function. v(1) = 0.04(1)^3 - 0.06(1)^2 = -0.02 ft/s.

(c) The particle is at rest when its velocity is equal to zero. To find the time(s) the particle is at rest, we set v(t) = 0 and solve for t. So, the particle is at rest at t = 0 and t = 1.5 seconds.

(d) The particle is moving in the positive direction when its velocity is positive. To find the time(s) the particle is moving in the positive direction, we look for the intervals where v(t) > 0. The particle is moving in the positive direction when 0 < t < 1.5 seconds.

(e) The total distance traveled during the first 12 seconds can be found by finding the area under the velocity-time graph. We integrate v(t) from 0 to 12 seconds. The total distance traveled during the first 12 seconds is approximately 96 ft.

(f) To find the acceleration at time t, we differentiate the velocity function. The derivative of v(t) = 0.04t^3 - 0.06t^2 is a(t) = 0.12t^2 - 0.12t ft/s^2.

(g) To find the acceleration after 1 second, we substitute t = 1 into the acceleration function. a(1) = 0.12(1)^2 - 0.12(1) = 0 ft/s^2.