Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

Answers

Answer 1
Answer:

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s


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Write a paragraph or two (no more than 1 page) describing the misconception and then explain the correct physics by identifying the forces or physics concepts involved in simple terms, like you are talking to your 3rd grade sibling. Start with a question (will water fall out of the bucket when I stop rotating it?)
Give the common answer that is incorrect (what does your intuition tell you should happen?).
Explain what happens and why in simple terms (try practicing on an elementary student to make sure they understand).
Use diagrams and sketches in your explanation and avoid using physics terms like rotational, centripetal, normal forces, tension, friction, acceleration, velocity, vectors, buoyancy, energy, kinetics, potential, heat capacity, etc.

Answers

Answer:

1.) Everything that moves, will eventually come to a stop. Rest is the “natural” state of all objects

Of all physics misconceptions, this is the most common. Even the great philosopher Aristotle, included it into his most important contribution to the field, his famous Laws of Motion. But now we know it is wrong because Newton’s First Law of Motion tells us that “everything at rest will stay at rest, and everything in motion will stay in motion, unless acted upon by an external force.”

The first statement seems reasonable enough, but the second part is a little bit murky. The reason this confusion persists boils down to the fact that we are unable to identify the force that stops all motion, which is friction. Friction is a force that acts between two objects that are in contact and are moving relative to each other. When we roll a ball, it stops because of the frictional force acting between it and the floor.

2.) A continuous force is needed for continuous motion

This misconception is a direct consequence of the first one. While this is true, if you are, for example, pushing a grocery cart in a supermarket, again this is only because there is friction involved. The force you apply to keep an object moving is only to counteract the frictional force. If you were to throw a rock on outer space, it would travel with a constant velocity forever, unless it hits something, of course. This is because space is mostly empty (it has trace elements of gas and dust throughout), and there would not be any frictional force acting on that rock.

3.) An object is hard to push because it is heavy

This is one of the most common misconceptions because it’s something we see and feel everyday. While a heavy object is really hard to push, it is not because of its weight, but because of its inertia or mass. Inertia is an objects resistance to change in motion. It is important to note that inertia is resistance to “change motion” rather than just motion itself. When, I was a kid, I imagined that it would be easy to carry and push massive objects when in outer space, but not surprisingly, my younger self was wrong.,

With that said… Since these objects still have mass despite being weightless, this mass represents the object’s inertia.

4.) Planets revolve around the sun because they are pushed by gravity

We have to remember that gravity — the weakest of the four fundamental forces — is an attractive force. The reason why planets revolve around the Sun can be chalked up to the fact that the planets were already spinning within the protoplanetary disk encircling a young Sun. Gravity merely keeps the planets in orbit around the Sun, but it isn’t necessarily the one thing pushing the planets along their orbital plane.

5.)  Heavier objects fall faster than lighter ones

This misconception is already debunked long ago by Galileo on his experiment when he dropped two objects with different masses on the Leaning Tower of Pisa. He has shown on that experiment that objects move downward with the same acceleration.

Again, the problem comes from not being able to identify another force that is involved, which is air resistance. All objects moving through air, and hence, all falling objects, experience air resistance. This force is proportional to the area of the object in the direction of motion. Usually, this force is negligible, but for light objects — with weight comparable to the air resistance, like a feather — it will have a big effect. This is ultimately confirmed by the famous hammer and feather drop experiment on the moon.

6.) There is no gravity in outer space

There is gravity in outer space, it is just weaker than what we experience here on Earth. Astronauts that are orbiting the Earth don’t experience gravity because they are free-falling (yes, you read that right). All satellites, including the moon and the planets, are in a constant state of freefall.

They just also have a tangential velocity with their free fall, that is why they don’t crash to what they are orbiting. When something is in free fall, it becomes weightless. This is why Kate Upton can do a photo shoot in zero gravity here on Earth. The plane that they are riding in actually went into free fall to do that.

7.) Planets move in circular orbits around the Sun

Planets actually move in elliptical orbits around the sun (with the Sun being the focus of the ellipse). This is actually the first of Kepler’s Three Laws of Planetary Motion, which deals with precisely how planets orbit the Sun.

One misconception deals with our seasons. Some might wrongly come to the conclusion that Earth’s proximity to the Sun dictates the seasons (summer is when Earth is closest to the Sun and winter is when it’s farther away), but that’s not entirely true. In reality, our seasons are caused by the tilt of Earth’s axis.

Much of our knowledge of the interior of the Earth comes from the study of planetary vibrations, which is the science of

Answers

Answer:

Seismology.

Explanation:

  • Seismology is the beach of physical science that deals with the study of vibrations that comes out from the interior of the earth onto the surface and these vibrations are in the form of seismic waves that are primary, secondary, and surface waves.
  • The science of seismology tells about the magnitude and intensity of these waves that lead to planetary vibrations. These waves trigger earthquakes, floods, and even landslides.

Energy is the capacity to do work, but not to produce heat

Answers

Energy and Work Energy is the capacity to do work or to produce heat. Internal energy is the sum of kinetic energy and the potential energy. ... The KE would increase because heating something causes an increase in temperature.

Final answer:

Energy is the capacity to do work but not to produce heat. In physics, energy can exist in various forms, including mechanical and thermal energy.

Explanation:

Energy is the capacity to do work and is an important concept in physics. In the context of this question, it is stated that energy is the capacity to do work but not to produce heat. This highlights the distinction between the two forms of energy. For example, mechanical energy can be used to perform work on an object and cause it to move, while thermal energy is associated with heat and not directly related to work. However, it's important to note that energy can be converted from one form to another, such as converting mechanical energy to thermal energy in a friction process.

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On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s) ?

Answers

Answer:

P=13.5 W

Explanation:

In this case, power is the rate of transferring heat per unit time:

P=(Q)/(\Delta t)(1)

The heat is given by the formula of the latent heat of fusion, since the ice is melting.

Q=mL_f(2)

Here m is the ice's mass and L_f is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

P=(mL_f)/(\Delta t)\nP=(3.5kg(334*10^3(J)/(kg)))/(86400s)\nP=13.5 W

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3. (a) Draw a force diagram for the balloon. (Submit a file with a maximum size of 1 MB.) (b) Calculate the buoyant force acting on the balloon. (Give your answer to at least three significant figures.) 4159 N (c) Find the net force on the balloon. 1524 N Determine whether the balloon will rise or fall after it is released. The balloon will (d) What maximum additional mass can the balloon support in equilibrium? 155 kg (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? The balloon and its load will remain stationary. The balloon and its load will accelerate downward. The balloon and its load will accelerate upward. (f) What limits the height to which the balloon can rise?

Answers

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

W=mg

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

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

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

(c) 1524 N

The mass of the helium gas inside the balloon is

m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

where \rho_h is the helium density; so we the total mass of the balloon+helium gas inside is

m=m_h+m_b=59 kg+210 kg=269 kg

So now we can find the weight of the balloon:

W=mg=(269 kg)(9.8 m/s^2)=2635 N

And so, the net force on the balloon is

F=B-W=4159 N-2635 N=1524 N

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

W'=(m'+m)g=B

where m' is the additional mass. Re-arranging the equation for m', we find

m'=(B)/(g)-m=(4159 N)/(9.8 m/s^2)-269 kg=155 kg

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

B=\rho_a V g

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

Final answer:

The physics involved in the functioning of helium balloons is based on buoyancy and Archimedes' Principle. The forces at play include the force due to gravity, the buoyant force and the net force, which determines the motion of the balloon. The balloon's height limit is determined by the decrease in air density with altitude.

Explanation:

The several parts of this question are related to the principles of buoyancy and Archimedes' Principle. First, regarding the force diagram for the balloon (part a), it would show two primary forces. The force due to gravity (Fg) acting downwards and the buoyant force (Fb) acting upwards, which is a result of the displacement of air by the balloon. The net force mentioned in part (c) is calculated as the difference between these two forces.

Calculating the buoyant force (part b) involves multiplying the volume of the balloon by the density of the air and the acceleration due to gravity (Fb = V * ρ_air * g). For the net force on the balloon (part c), this is calculated by subtracting the weight of the balloon from the buoyant force (F_net = Fb - Fg). If the net force is positive, the balloon will rise, if it's negative, the balloon will fall, and if it is zero, the balloon will remain stationary.

The maximum additional mass the balloon can support in equilibrium (part d) is calculated using the net force divided by gravity. If the mass of the load is less than this value (part e), the balloon and its load will accelerate upward.

Lastly, the limit to the height to which the balloon can rise (part f) is determined by the decreasing density of the air as the balloon ascends. The buoyant force reduces as the balloon rises because the air density is lower at higher altitudes.

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Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answers

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

Final answer:

The magnetic field at point P, which is equidistant from two long parallel wires with equal anti-parallel currents, is calculated using Ampere's law. The net magnetic field is zero because the fields due to each wire cancel each other at that point.

Explanation:

The question concerns the calculation of the magnetic field at a point equidistant from two long parallel wires that carry equal anti-parallel currents. According to the right-hand rule and Ampere's law, each wire generates a magnetic field that circles the wire. For two wires carrying currents in opposite directions, the magnetic fields at the midpoint between the wires will point in opposite directions, thus they will subtract from each other when calculating the total magnetic field at point P.

To find the magnetic field at point P, we use the formula for the magnetic field due to a long straight current-carrying wire: B = (μ₀I)/(2πd), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10-7 T·m/A), I is the current, and d is the distance to the point of interest from the wire. In this case, the distance d will be the radius R = 5.00 cm since point P is equidistant from both wires.

Substituting the values into the formula, the magnetic field due to each wire at point P can be calculated. However, since the currents are anti-parallel, the net magnetic field at P would be the difference between the two fields, which is zero.