The process of cutting lines of magnetic flux with a conductor, and generating voltage, is the basisi of alternator and generator operation.True / False.

Answer:

The Scenario:

On a normal Sunday afternoon Mr. Golanski is sitting in his living room reading his book. He decides after a while to turn on the Television to see what’s on the news, (Mr. Golanski is using Radio waves when he turns his television on by signaling the TV from his remote control). After a few hours Mr. Golanski decides it’s time to have dinner. He heats up a quick meal in his microwave because he doesn’t have the patience for cooking. (He is using microwave radiation to heat his food because water molecules in food absorb the radiation). He sits down for his meal, and halfway through he starts to choke! In a panicked frenzy he runs to his bathroom to try and dislodge the obstacle from his throat. By doing so he switched on the fluorescent lights in his bathroom exposing himself to small amounts of ultraviolet radiation. (Fluorescent lights absorb UV radiation and transmit visible light along with small amounts of UV light). Unable to dislodge the obstacle from his throat Mr. Golanski seeks help from his neighbor who drives him straight to the ER. To treat him properly the physicians opt for a fluoroscopy to examine Mr. Golanski’s esophageal tract. (Thus he is making use of X-ray imaging to obtain a visual of his internal esophageal structure to check for the obstruction). Once treated and discharged from the hospital Mr. Golanski returns home grateful to have survived this ordeal with minimum damage.

Explanation:

The Electromagnetic Spectrum is the range of frequencies and wavelengths for different light waves. They range with increasing frequency from Radio waves, to Microwaves, to Infrared waves, to Visible waves, to Ultraviolet waves, to Infrared waves, to X-rays, and to Gamma rays. Several of which we use in our daily lives such as Radio waves when operating our television or using our cellular phones. We also use microwaves to heat our food or for communication with satellites. We are also exposed to natural Ultraviolet radiation from the sun; however, we can also get exposed to other forms such as from certain types of light bulbs. We see visible light in the form of all the colors we can detect around us. We make use of x-rays for imaging techniques widely used in medicine for diagnostics, as well as Infrared waves in our home security systems. The electromagnetic spectrum is always used as a part of our everyday life.

Answer:

15 MPH or less than that

Explanation:

You shouldn't drive faster than 15 MPH so if a truck unexpectedly pulls out you'll have time to stop.

When parked cars, branches of the tree, fences, houses, or other things block your vision of an open intersection, you can pause before approaching the junction and move slowly ahead until you can see if there is cross-traffic before moving through the junction.

Answer:

Percolation of water through the soil.

Explanation:

When a tray containing a mixture of dough and sand is propped on one end of the tray with bricks then there is a slope from one end of the tray to the other end of the tray.

When water is poured on the higher end of the tray then the it tries to flow from the higher altitude to the lower altitude of the tray by the process of percolation through the granular spaces created due to the mixing of sand in the dough. Sand being granular in structure is easily permeable to water and passes water easily through its tiny pores.

Similar is the process of percolation of water through the layers of soil on the earth which also contains granular particles in them.

Answer: 3.02kg

Explanation:

According to the law of conservation of momentum, the sum of momentum of bodies before collision is equal to the sum of momentum of the bodies after collision.

Note that this bodies will move with a common velocity after collision.

Since momentum = mass of a body × its velocity

Let m1 and m2 be the masses of the spheres

u1 and u2 be their velocities

v be their common velocity after collision

Mathematically

m1u1 + m2u2 = (m1+m2)v

From the question, the second sphere is initially at rest i.e u2 = 0m/s and the composite system moves with a speed equal to one-third the original speed of 1st sphere i.e V = u1/3

Substituting this conditions into the formula, we have;

m1u1 + m2(0) = (m1+m2)u1/3

m1u1 = u1/3(m1+m2)

Given m1 = 1.51kg

m1 = 1/3(m1+m2)

m1 = m1/3 + m2/3

m1-m1/3 = m2/3

2m1/3 = m2/3

2m1 = m2

2(1.51) =m2

m2 = 3.02kg

The mass of the second sphere is 3.02kg

Answer:

A=0.1024 m

Explanation:

Given Data

M=3.60 kg

m=42.0 g=0.042 kg

v₀=140 m/s

k=560 N/m

To find

Amplitude of Simple Harmonic Motion

Solution

Use conservation of momentum to solve for velocity

[tex]mv=(m+M)V\\V=\frac{mv}{m+M}\\ V=\frac{0.042kg*140m/s}{0.042kg+3.60kg}\\ V=1.6145 m/s[/tex]

Use conservation of energy to solve for amplitude

ΔKE+ΔUg+ΔUs=0

ΔKE+0+ΔUs=0

1/2MV²-1/2kA²=0

[tex]A=\sqrt{\frac{MV^{2} }{k} }\\ A=\sqrt{\frac{(3.642kg)*(1.6145m/s)}{560N/m} }\\ A=0.1024m[/tex]

Answer:

electric force of attraction, The work  is zero

Explanation:

Between a positive and a negative charge there is an electric force of attraction, as in this case the particle moves in an orbit circulates at a certain velocity at an angle between the displacement and the force is 90, consequently the work defined by

                   W = F d cos θ

The work  is zero, therefore the elective force does not create work on the load.

Both scenarios involve work done by forces - the electric force in the first scenario causes a charged particle to move, changing its potential energy. In the second scenario, a magnetic force does work on an iron nail, resulting in its displacement.

In the provided scenarios, there are two primary forces at play which are the electric force and the magnetic force.

The first scenario considers a positively charged particle moving in a circle at a fixed distance from a negatively charged particle. Here, the force present is an electric force. The electric force causes the positively charged particle to move in a direction of the force field lines. As a result of the force field, work is indeed done. This is because the electrical field does work on the charge, altering its potential energy.

In the second scenario, an iron nail is being pulled from a magnet. Here, the force is magnetic. The movement or displacement of the nail from the magnet is a clear indication that work has been done, since work is done when a force moves an object. The force exerted by the magnet does work on the nail by overcoming the magnetic force that initially held the nail in place.

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Answer:

Work energy theorem: [tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]

Conservation of energy:

This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.

[tex]PE=m.g.h[/tex]

Explanation:

Work energy theorem:

It states that the total work done by the sum of all the forces acting on a particle is equal to the change in kinetic energy of the particle.

Mathematically:

[tex]W=\Delta KE=\frac{1}{2} m.v_f^2-\frac{1}{2} m.v_i^2[/tex]

Conservation of energy:

This theorem states that energy can neither be created nor be destroyed it can only get converted from one form to another.

Relation between the Forces and potential energy:

Force here implies the force of gravitation. Potential energy in general means the energy which is contained in a body due to virtue of its position in a field of influence.

Here the field of influence is gravity. So,

[tex]PE=m.g.h[/tex]

where:]

m = mass of the body

g = acceleration due to gravity

h = height of the object from the earth surface

Answer:

Explanation: Peripherial arterial disease is a blood circulation disorder which occurs when the blood vessels outside the heart and brain to block or spasm. This is as a result of the pulse decreasing rather than increasing in amplitude.

Answer:

θ=5.65°

Explanation:

Given Data

Mass m=1.5 kg

Length L=0.80 m

First spring constant k₁=35 N/m

Second spring constant k₂=56 N/m

To find

Angle θ

Solution

As the both springs take half load so apply Hooks Law:

Force= Spring Constant ×Spring stretch

F=kx

x=F/k

as

[tex]d=x_{1}-x_{2}\\ as \\x=F/k\\so\\d=\frac{F_{1} }{k_{1}} -\frac{F_{2}}{k_{2}}\\ Where \\F=1/2mg\\d=\frac{(1/2)mg}{k_{1}} -\frac{(1/2)mg}{k_{2}}\\ d=\frac{mg}{2}(\frac{1}{k_{1}} -\frac{1}{k_{2}} )\\ And\\Sin\alpha=d/L\\\\alpha =sin^{-1}[\frac{mg}{2L}(1/k_{1}-1/k_{2})]\\\alpha =sin^{-1}[\frac{(1.5kg)(9.8m/s^{2} )}{2(0.80m)}(1/35Nm-1/56Nm) ]\\\alpha =5.65^{o}[/tex]

θ=5.65°

Answer:

46.67 N/m

Explanation:

mass, m = 0.1 kg

distance, y = 2.1 cm = 0.021 m

Let K be the spring constant.

F = mg = Ky

0.1 x 9.8 = K x 0.021

K = 46.67 N/m

Thus, the spring constant of the spring is 46.67 N/m.

Answer:

k = 46.67 N/m

Explanation:

given,

mass, m = 0.10 Kg

Length, x = 2.1 cm = 0.021 m

spring constant, k = ?

we know,

spring force

F = k x

and also

F = m g

computing both the equation

k x = m g

[tex]k = \dfrac{mg}{x}[/tex]

[tex]k = \dfrac{0.1\times 9.8}{0.021}[/tex]

k = 46.67 N/m

Spring constant for the spring is equal to k = 46.67 N/m

Answer:

M = 9.71 x 10³⁰ Kg

Explanation:

given,

Time period of the planet ,T = 3.31 x 10⁷ s

distance from the star, r = 2.62 x 10¹¹ m

mass of the star = ?

using Kepler third law

[tex]T^2 = \dfrac{4\pi^2}{GM}r^3[/tex]

now,

[tex]M = \dfrac{4\pi^2}{GT^2}r^3[/tex]

where as, G = 6.67 x 10⁻¹¹ m³/kg.s²

[tex]M = \dfrac{4\pi^2}{6.67\times 10^{-11}\times (3.31\times 10^7)^2}\times (2.62\times 10^{11})^3[/tex]

M = 9.71 x 10³⁰ Kg

The mass of the star is equal to M = 9.71 x 10³⁰ Kg.

Kepler's third law can be used to calculate the mass of a star from the orbit of a planet. Given the values for the period and distance of the planet from the star, the calculation results in an estimated mass of 1.2 times the mass of the Sun.

To calculate the mass of a star from the orbit of a planet, we need to use Kepler's third law as reformulated by Newton. The equation is D³ = (M₁ + M₂)P² , where D is the semi-major axis (distance of the planet from the star), M₁ is the mass of the star, M₂ is the mass of the planet, and P is the period of the planet's orbit. In this case, because the mass of the planet is so small compared to the star, we can ignore M₂ and rearrange the formula to solve for M₁: M₁ = D³ / P² .

To plug in the values given, first, take note that D is given in meters and P in seconds. So first we have to convert D from meters to astronomical units (AU) since 1 AU = 1.496 × 10^11 m. So, D = 2.62×10^11 m / 1.496 × 10^11 m/AU = 1.75 AU.

Now we substitute these in our equation, we have M₁ = (1.75)³ AU³ / (3.31 × 10^7 s)² = 1.2 times the mass of the Sun (since the unit of star mass in Kepler's law is a solar mass).

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statement accurately describes something that Yolanda can do as a part of her study She can use destructive interference to generate a wave with an amplitude of 2.8 m.

The correct option is (c).

To determine the resulting amplitude of waves under interference, we consider the principles of constructive and destructive interference.

1. Constructive Interference: When two waves meet crest to crest or trough to trough, they add up to produce a wave with an amplitude equal to the sum of the individual amplitudes.

  For Yolanda's waves:

[tex]\[ \text{Amplitude} = 2 \, \text{m} + 3 \, \text{m} = 5 \, \text{m} \][/tex]

  Option b. She can use constructive interference to generate a wave with an amplitude of 1.5 m is incorrect because the resulting amplitude from constructive interference must be greater than or equal to the sum of the individual amplitudes.

2. Destructive Interference: When two waves meet crest to trough, they cancel each other out partially or completely. The resulting amplitude is the absolute difference between the amplitudes of the individual waves.

  For Yolanda's waves:

[tex]\[ \text{Amplitude} = |2 \, \text{m} - 3 \, \text{m}| = |-1 \, \text{m}| = 1 \, \text{m} \][/tex]

  Option a. She can use destructive interference to generate a wave with an amplitude of 3.1 m is incorrect because the resulting amplitude from destructive interference must be less than or equal to the difference between the individual amplitudes.

Option d. She can use constructive interference to generate a wave with an amplitude of 3.5 m is incorrect because the resulting amplitude from constructive interference must be greater than or equal to the sum of the individual amplitudes.

Therefore, option c. She can use destructive interference to generate a wave with an amplitude of 2.8 m is the accurate statement.

Answer:

Would life be different if the electron were positively charged and the proton were negatively charged?

A) Yes.

B) No.

Answer is NO (Option B)

Does the choice of signs have any bearing on physical and chemical interactions?

A) Yes.

B) No.

Answer is NO (Option B)

Explanation:

Electric charges are of two general types: positive and negative, electrons are designated to be the carriers of negative charge. The charge on the proton and electron are exactly the same size but opposite. Opposite charges would still attract, and like charges would still repel in other words electrons would still repel other electrons, but protons and electrons would still attract each other. The designation of charges as positive and negative is merely a definition.

The choice of signs have no bearing on physical and chemical interactions because, proton and electron is just a name assigned to the sign as like charges repel and unlike charges attract. An atom is built with a combination of three distinct particles: electrons, protons, and neutrons. Electric charge, which can be positive or negative,  is neither created nor destroyed. The atom that has lost its equal share in the bonding electron pair acquires a partial positive charge because its nuclear charge is no longer fully canceled by its electrons.

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 11.9 m/s²  

Final velocity, v = ?

Displacement,s = 400 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 11.9 x 400

v = 97.57 m /s

Speed of the race car at the end of the run is 97.57 m/s

Answer:

[tex]\displaystyle \theta' =0.24\ rad/s[/tex]

Explanation:

Rate Of Change

Let some variable y depend on time t. we can express y as a function of t as

[tex]y=f(t)[/tex]

The instant rate of change of y respect to t is the first derivative, i.e.

[tex]y'=f'(t)[/tex]

The balloon, the ground and the observer form a right triangle (shown below) where the height of the balloon y, the horizontal distance x, and the angle of elevation are related with the trigonometric formula

[tex]\displaystyle tan\theta =\frac{y}{x}[/tex]

Since x is constant, we take the derivative with respect to time  by using the chain rule:

[tex]\displaystyle sec^2\theta \ \theta' =\frac{y'}{x}[/tex]

Solving for [tex]\theta'[/tex]

[tex]\displaystyle \theta' =\frac{y'}{xsec^2\theta}[/tex]

Let's compute the actual angle with the initial conditions y=9 feet, x=12 feet

[tex]\displaystyle tan\theta =\frac{y}{x}[/tex]

[tex]\displaystyle tan\theta =\frac{9}{12}=\frac{3}{4}[/tex]

Knowing that

[tex]\sec^2\theta=1+tan^2\theta[/tex]

[tex]\displaystyle \sec^2\theta=1+\left(\frac{3}{4}\right)^2[/tex]

[tex]\displaystyle \sec^2\theta=\frac{25}{16}[/tex]

The balloon is rising at y'=8 feet/sec, thus we compute the change of the angle of elevation:

[tex]\displaystyle \theta' =\frac{8}{12\ \frac{25}{16}}[/tex]

[tex]\displaystyle \theta' =\frac{32}{75}\ rad/s[/tex]

[tex]\boxed{\displaystyle \theta' =0.43\ rad/s}[/tex]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

[tex]s_1 = v_1t = 1.6t[/tex]

Similarly the distance (m) from your friend building to the restaurant:

[tex]s_2 = v_2t = 1.2t[/tex]

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

[tex]s_1 + s_2 = 5b[/tex]

[tex]1.6t + 1.2t = 5b[/tex]

[tex]2.8t = 5b[/tex]

[tex]t = 5b/2.8 = 25b/14[/tex]

[tex]s_2 = 1.2t = 1.2(25b/14) = 2.143b[/tex]

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

Answer:

-[tex]29.2\times 10^{3} N[/tex]

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time=[tex]\Delta t[/tex]=1.3 s

We have to find the average force exerted on the car.

Average force=[tex]\frac{change\;in\;momentum}{\Delta t}[/tex]

[tex]F_{avg}=\frac{mv-mu}{1.3}[/tex]

[tex]F_{avg}=\frac{1900(0)-1900(20)}{1.3}[/tex]

[tex]F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N[/tex]

Hence, the average force exerted on the car that hits a line of water barrels=-[tex]29.2\times 10^{3} N[/tex]

Final answer:

Using the change in velocity and time, the acceleration of the car is calculated, followed by the application of Newton's second law to find the average force exerted on the car as 29,222 N.

Explanation:

To determine the average force exerted on the car that hits a line of water barrels, we can use the equations of motion and Newton's second law of motion. The car goes from 20 m/s to rest, which means that its change in velocity is -20 m/s. Given the time of 1.3 seconds for this change, we can find the acceleration using the formula:

a = Δv / t

Where Δv is the change in velocity and t is the time taken. The acceleration a is therefore:

a = -20 m/s / 1.3 s = -15.38 m/s²

Now, using Newton's second law of motion, Force (F) = mass (m) × acceleration (a), we can find the force:

F = 1900 kg × -15.38 m/s² = -29222 N

The negative sign indicates that the force is in the opposite direction to the motion of the car. Thus, the average force exerted on the car is 29,222 N.

Final answer:

Using the change in velocity and time, the acceleration of the car is calculated, followed by the application of Newton's second law to find the average force exerted on the car as 29,222 N.

Explanation:

To determine the average force exerted on the car that hits a line of water barrels, we can use the equations of motion and Newton's second law of motion. The car goes from 20 m/s to rest, which means that its change in velocity is -20 m/s. Given the time of 1.3 seconds for this change, we can find the acceleration using the formula:

a = Δv / t

Where Δv is the change in velocity and t is the time taken. The acceleration a is therefore:

a = -20 m/s / 1.3 s = -15.38 m/s²

Now, using Newton's second law of motion, Force (F) = mass (m) × acceleration (a), we can find the force:

F = 1900 kg × -15.38 m/s² = -29222 N

The negative sign indicates that the force is in the opposite direction to the motion of the car. Thus, the average force exerted on the car is 29,222 N.

Answer:

3035281543986.976 mi

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = g

Time taken

[tex]t=1\ y\\\Rightarrow t=365.25\times 24\times 60\times 60[/tex]

u = Initial velocity = 0

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times (365.25\times 24\times 60\times 60)^2\\\Rightarrow s=4.8848\times 10^{15}\ m[/tex]

Converting to mi

[tex]1\ m=\dfrac{1}{1609.34}\ mi[/tex]

[tex]4.8848\times 10^{15}\ m=4.8848\times 10^{15}\times \dfrac{1}{1609.34}\ mi\\ =3035281543986.976\ mi[/tex]

Number of miles travelled in one year is 3035281543986.976 mi

Answer:

b

Explanation:

Final answer:

The correct term for the physical act of moving air into and out of the lungs is pulmonary ventilation, which is option d in the provided choices.

Explanation:

The physical act of moving air into and out of the lungs is called ventilation. This process is part of respiration, which is much more than just breathing. Respiration also includes the exchange of gases between the blood and the cells of the body, not just the action of breathing itself. Breathing, or ventilation, involves two primary steps: inhaling, which is an active process driven by the contraction of the diaphragm, and exhaling, which is typically a passive process caused by the lungs' elasticity when the diaphragm relaxes.

Pulmonary ventilation is the term that accurately describes the act of breathing, involving the precise movement of air into and out of the lungs. Therefore, the correct answer to the question is option d. pulmonary ventilation.

Answer:

True

Explanation:

True

Diffusion refers to that process through which concentration is transmitted from one place to another concentration i.e from higher concentration to lower concentration. Due to this only, characteristics able to spread all over the space.

it can only happen in fluid(liquid and gas) because their particle can move in a random direction

Answer:

3.14 m/s^2 is the acceleration

Explanation:

You must use the formula Fnet=ma

So the total forces acting on the system are the 450N push force and the 35N friction force, which acts in the opposite direction.

Thus, the sum of the forces is 450 -35= 415N

Also you know the mass is 132 kg, so to find the acceleration divide the previously calculated net force by the mass

So 415/132 =3.14 m/s^2

Hope this helped:)

The bobsled with a mass of 132 kg and a net force of 415 N after factoring in friction accelerates at about 3.14 meters per second squared.

To determine the acceleration of the bobsled, we will use Newton's second law, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. The net force in this scenario is the combined push minus the force of friction. Therefore, the equation to use is:

Net force = Total push - Force of friction

Net force = 450.0 N - 35 N = 415 N

Now, we use this net force to find the acceleration:

Acceleration (a) = Net force / Mass

a = 415 N / 132 kg

a ≈ 3.14 m/s2

The bobsled will accelerate at approximately 3.14 meters per second squared when the net force acting on it is 415 N.

Answer:

46%  (0.46)

Explanation:

temperature of hot reservoir (Th) = 1.3 kJ

temperature of COLD reservoir (Tc) = 0.7 kJ

Efficiency = 1 - (Tc/Th)

Efficiency = 1 - (0.7/1.3) = 0.46 = 0.46 x 100 = 46 %

Answer:

46.2 %

Explanation:

heat in (Q_in) = 1.3kJ

heat out (Q_out) = 0.7kJ

work out = heat in - heat out = 1.3 - 0.7 = 0.6 kJ

efficiency = work out / heat in * 100%

                = 0.6/1.3 * 100 % = 46.2 %

The man's centre of mass will shift upward when he raises his arms from down to up. However, the exact amount of shift for the man plus his arms as a system would need more detailed information and would involve calculations including both the body's and the arms' centres of mass.

The subject matter of this question is high school-level physics, specifically, mechanics and the concepts of center of mass and system dynamics. The question asks, 'When a 70 kg man raises both his arms, from hanging down to straight up, by how much does he raise his centre of mass?' To answer this, we need to consider the centre of mass of the person's body separately and the arms as separate objects as well.

Let's take the arm's centre of mass to be at its mid-point due to it being modelled as a uniform rod. When the arms are hanging down, the centre of mass of each arm is 75 cm from the ground. When the arms are raised straight up, the centre of mass of each arm is moved to a height equivalent to the man's height plus 75/2 cm.

However, to get the overall shift in the centre of mass of the man plus his arms as a whole system, we would need more specific details about the man's body proportions. Generally, the human body's centre of mass with arms hanging down by the sides is around 56% of the person's height from the ground. When the arms are raised, the body's individual centre of mass doesn't shift much, but the overall system's centre of mass (of the man and his arms together) does shift upward, though how much exactly requires more detailed anthropomorphic data.

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When a 70 kg man raises both his arms from hanging down to straight up, his center of mass raises by 0.0375 meters (3.75 cm).

To determine how much a 70 kg man raises his center of mass when he raises both arms from hanging down to straight up, we model his arms as uniform rods. Given:

Length of each arm ( L) = 75 cm = 0.75 m

Mass of each arm ( M) = 3.5 kg

Total mass of the man ( M_{total}) = 70 kg

The center of mass of each arm when it is hanging down is at half its length, i.e., 0.75 m / 2 = 0.375 m from the shoulder. When the arms are raised straight up, the center of mass of each arm moves to a point 0.75 m above the shoulder.

The vertical displacement ( Δh) of the center of mass of each arm is thus:

Δh = 0.75 m - 0.375 m = 0.375 m

To find the overall shift in the man's center of mass, we calculate the contribution of the arms' displacement to the man's center of mass:

The change in center of mass ( ΔCM) is given by:

ΔCM = (2 x M x Δh) / M_{total}

Substituting the values, we get:

ΔCM = (2 x 3.5 kg x 0.375 m) / 70 kg = 0.0375 m

Thus, the man's center of mass raises by 0.0375 meters (3.75 cm).

Answer:

d. supply of the most limited resources.

Explanation:

the carrying capacity of a species in an environment is determined by the number of individual in an environment that the limited resources can sustain indefinitely without fighting for food, vegetation, water and light to sustained themselves.

the carrying capacity can be determined by how much the limited resources in the environment can sustain the species. the resources needed by species in a particular environment varies from habitat to habitat.

for some species, the limiting resources needed is food which consequently affect the way they reproduced and affect the population size.

The carrying capacity of an environment for a specific species is defined by the availability of the most limited resources, and it indicates the maximum population size that the environment can sustain.

The carrying capacity of an environment for a particular species at a particular time is determined by the supply of the most limited resources. This refers to the maximum population size of a species that an environment can sustain indefinitely, given the availability of resources such as food, water, and space. It's not solely determined by the number of individuals in the species, the distribution of the population, or the reproductive potential of the species, rather it is driven by the resource which is in shortest supply.

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Final answer:

The actual distance from Albuquerque to Santa Fe is 90 km.

Explanation:

To find the actual distance between Albuquerque and Santa Fe, we first need to find the scale factor. The scale on the map tells us that 1 cm represents 45 km. Since the distance on the map is 2 cm, we can multiply it by the scale factor to get the actual distance: 2 cm * 45 km/cm = 90 km.

Therefore, the actual distance from Albuquerque to Santa Fe is 90 km, so the correct answer is option A.

Answer:

<-0.45, 0, 0> kgm/s

Explanation:

The change in momentum would equal to the impulse generated by force F = <-0.3, 0, 0> over time duration of Δt = 1.5 second. The impulse is defined as the product of force F and the duration Δt

[tex]\Delta p = F \Delta t = (-0.3) * 1.5 = -0.45 kgm/s[/tex]

So the change in momentum is <-0.45, 0, 0> kgm/s

Final answer:

The change in momentum of the fancart is < -0.45, 0, 0 > kgm/s, which is calculated using the impulse-momentum theorem by multiplying the force applied by the time interval over which it was applied.

Explanation:

The question pertains to change in momentum of a fancart when a constant force is applied to it. The change in momentum (delta p) can be calculated using the impulse-momentum theorem, which states that the change in momentum is equal to the impulse applied to the object. The impulse is the product of the force and the time over which the force is applied. Given a force of < -0.3, 0, 0 > N applied for 1.5 seconds to a cart with an initial mass of 0.8 kg and initial velocity of < 0.9, 0, 0 > m/s, we can calculate the change in momentum:

Impulse = Force times Time

Impulse = < -0.3, 0, 0 > N
times 1.5 s = < -0.45, 0, 0 > Ns

Since delta p = Impulse, the change in momentum is also < -0.45, 0, 0 > kgm/s.

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

[tex]KE_r = \dfrac{1}{2}I\omega^2[/tex]

I of the moment of inertia of the sphere

[tex]I = \dfrac{2}{5}MR^2[/tex]

 v = R ω

using conservation of energy

[tex]KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2[/tex]

[tex]KE_r = \dfrac{1}{5}MV^2[/tex]

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

[tex]\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h[/tex]

[tex] 0.7 V^2 = 0.7 V'^2 + gh[/tex]

[tex] 0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53[/tex]

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

The linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

The given parameters;

Apply the principle of conservation of energy to determine the linear speed of the ball when it reaches the top of the ramp.

[tex]K.E_i + P.E_i = K.E_f + P.E_f \\\\(K.E + K.E_r)_i + P.E_i = (K.E + K.E_r)_f \ + P.E_f\\\\(\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_i + 0 = (\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_f + mgh[/tex]

where;

[tex](\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} mR^2\times \frac{v^2}{R^2} )_i = (\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} MR^2\times \frac{v^2}{R^2} )_f + mgh\\\\(\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_i = (\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_f + mgh\\\\0.7 mv_i^2 = 0.7mv_f^2+ mgh\\\\0.7 v_i^2 = 0.7v_f^2+ gh\\\\0.7v_f^2 = 0.7v_i^2 -gh\\\\v_f= \sqrt{\frac{0.7v_i^2 -gh}{0.7} } \\\\v_f = \sqrt{\frac{0.7(2.85)^2 -9.8(0.53)}{0.7} }\\\\v_f = 0.84 \ m/s[/tex]

Thus, the linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

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Final answer:

To calculate the spring force constant k in this problem, we can use Hooke's Law. By substituting the given values, we can calculate the spring force constant as approximately 124.97 N/m.

Explanation:

To calculate the spring force constant k, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, the ball compresses the spring to a maximum displacement of 4.68317 cm, so we can use this value in our calculation.

Hooke's Law can be written as F = k × x, where F is the force exerted by the spring, k is the spring force constant, and x is the displacement of the spring.

Given that the mass of the ball is 60.3 g, the acceleration of gravity is 9.8 m/s², and the displacement of the spring is 4.68317 cm, we can calculate the force exerted by the ball using the equation F = m × g, where m is the mass of the ball and g is the acceleration of gravity.

Substituting the values, we have F = (60.3 g) × (9.8 m/s²). Now, we can solve for k by dividing both sides of the equation F = k × x by x. This gives us k = F / x.

Substituting the values, we have k = (60.3 g × 9.8 m/s²) / (4.68317 cm).

Converting the values to SI units, we have k = (0.0603 kg × 9.8 m/s²) / (0.0468317 m).

Calculating k, we find that the spring force constant is approximately 124.97 N/m.

The spring force constant k,  is approximately 289.6 N/m.

To find the spring force constant k, we use the principle of conservation of energy. Here’s the step-by-step solution:

First, convert the given values to meters and kilograms:
Mass of the ball, m = 60.3 g = 0.0603 kg
Height from which the ball is dropped, h = 53.7 cm = 0.537 m
Maximum spring displacement, x = 4.68317 cm = 0.0468317 m

Calculate the potential energy at the initial height:
Potential Energy, PE = m × g × h
PE = 0.0603 kg × 9.8 m/s² × 0.537 m = 0.317 J

When the ball compresses the spring, all the potential energy converts into the elastic potential energy of the spring:

Elastic Potential Energy, EPE = (1/2) × k × x²
0.317 J = (1/2) × k × (0.0468317 m)²

Solve for k:
k = 2 × 0.317 J / (0.0468317 m)²
k ≈ 289.6 N/m

Therefore, the spring force constant k is approximately 289.6 N/m.

Answer:

[tex]\frac{k_eQ}{2h}[/tex]

Explanation:

The question is missing an image. I have added this as an attachment to my answer.

Given;

Q = total charge

R = radius of cylindrical shell

h = height of cylindrical shell

d = distance of point from the right side of the cylinder

Let the thickness of the cylindrical shell be [tex]dx[/tex] , and the charge  [tex]\frac{Qdx}{h}[/tex],

Now, using the formula for finding the electric field due to a ring at a chosen point:

[tex]dE = \frac{k_ex}{(x^2 + R^2)^{\frac{3}{2}}} {\frac{Qdx}{h}i}[/tex]

where [tex]x[/tex] = center of the ring to the point

[tex]k_e[/tex] = electrostatic constant

We integrate on both sides from the limits [tex]d[/tex] to [tex]d + h[/tex]  in order to determine the electric field at the point [tex]E[/tex]

[tex]\int\limits dE[/tex] = [tex]\int\limits^{d + h}_d {\frac{k_eQxdx}{h(x^2 + R^2)^{\frac{3}{2}}}i}[/tex]

[tex]E = \int\limits^{d + h}_d {\frac{k_eQdx}{h(x^2 + R^2)^{\frac{3}{2}}}i} = \frac{k_eQ}{2h}[/tex]  

The electric field at a point a distance d from the right side of the cylinder is :   E = [tex]\frac{K_{e}Q }{2h}[/tex]

Let's' assume thickness of cylindrical shell = dx and charge = [tex]\frac{Qdx}{h}[/tex]  

Next step : calculate the electric field due to a ring at a point

dE = [tex]\frac{kex}{(x^2+ R^2)^{\frac{3}{2} } } \frac{Qdx}{h} i[/tex]  --- ( 1 )

where : x = centre of the ring to the specific point

           ke = electrostatic constant

To determine the electric field constant by integrating equation within limits d to d + h

E = [tex]\int\limits^d_d {\frac{keQdx}{h(x^{2}+R^2)^{\frac{3}{2} } } } \, i = \frac{KeQ}{2h}[/tex]

Hence we can conclude that The electric field at a point a distance d from the right side of the cylinder is :   E = [tex]\frac{K_{e}Q }{2h}[/tex].

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