Newton's second law of motion tells us that the net force applied to an object equals its ________.

Dorsiflexion, as the word suggests, is the action of bending the foot upward at the ankle. In activities like walking, running, or ascending stairs, it is a typical movement. Plantar flexion is the term for the opposite motion.

Dorsiflexion is the name given to the process of bending the foot upward at the ankle. This is a typical movement when doing a variety of activities where your foot's position changes in relation to your leg, like walking, running, or ascending stairs. Plantar flexion is the term for the motion that pushes your foot downward in the opposite direction.

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Dorsiflexion refers to bending the foot upward at the ankle. This is one of the two primary movements at the ankle joint, the other being plantar flexion which involves lifting the heel or pointing the toes downward.

The term that refers to bending the foot upward at the ankle is called dorsiflexion. This movement is characterized by lifting the front of the foot so that the top of the foot moves toward the anterior leg. Dorsiflexion is one of two primary movements at the ankle joint. The opposite movement to dorsiflexion is plantar flexion, which involves lifting the heel of the foot from the ground or pointing the toes downward. These two movements, dorsiflexion and plantar flexion, are the only movements available at the ankle joint.

Dorsiflexion is the term that means to bend the foot upward at the ankle. It is a movement at the ankle joint where the top of the foot moves towards the anterior leg. This movement is opposite to plantar flexion, which is the bending of the foot downward at the ankle.

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Answer:

Theories may be revised over time as information is discovered or technology is developed.

Explanation:

A scientific theory is a set of knowledge related to a particular problem of interest, be it academic or practical. The structure of a theory is formed by the following main elements: problem, evidence, postulates, axioms, questions, hypotheses, predictions, theses, rules and laws. Theory seeks to explain a particular phenomenon or problem of society, so theories can be revised and change as society changes and new technologies and new information are discovered.

Answer: The Answer is Plasma

Explanation:

Answer:

C. Plasma

Explanation:

it was blue he vented

The gas is chlorine gas

Density at STP = 3.16 g per L

1 mole of a gas at STP occupies, 22.4 L

Therefore, it means mass of the gas in 22.4 L that is 1 mol gas will be;

= 3.16 g × 22.4 L

= 70.8 g/mol

≈ 71 g/mol

The identity of the gas would be Chlorine gas, whose molar mass is 71 g/mol

Given the density of the gas at standard temperature and pressure, we can use the molar volume of a gas at STP to determine the molar mass. Using this process, the gas is likely to be Bromine (Br2).

The density given is a key factor to determine the gas. Because the question refers to standard temperature and pressure (stp), we can use the known molar volume of a gas at stp, which is 22.4 L/mol. Based on the given density and the molar volume, we can calculate the molar mass of the unknown gas:

Density = Molar mass / molar volume,

so Molar mass = Density x molar volume = 3.16 g/L x 22.4 L/mol = 70.784 g/mol.

Based on common molar masses, this results indicates that the unknown gas is most likely Bromine (Br2), as its molar mass is approximately 79.9 g/mol, the closest common molar mass to 70.784 g/mol.

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The speed of the water will be "43.75 ft/sec".

According to the question,

Initially inner diameter,

Final diameter,

By continuity equation, we get

→ [tex]Area1\times Velocity1=Area2\times Velocity2[/tex]

By substituting the values, we get

→ [tex]pi\times (\frac{0.052}{2} )^2\times 7= pi\times (\frac{0.0208}{2} )^2\times Velocity[/tex]

→             [tex]Velocity = 43.75 \ ft/sec[/tex] (speed)

Thus the above answer is correct.

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The time taken for an object to hit the ground is independent of its mass when air resistance is neglected. Therefore, the second object will also take 9 seconds to hit the ground.

Understanding the Effect of Mass on Free Fall

When you drop an object, its time to reach the ground is determined by the laws of physics related to free fall. Under the influence of gravity, all objects fall at the same rate regardless of their mass, assuming we are neglecting air resistance.

Given: Mass of first object, [tex]m_1 = 5 kg[/tex], Time to hit the ground, [tex]t_1 = 9 s.[/tex]

The second object has a mass, [tex]m_2 = 10 kg[/tex]. To find the time [tex]t_2[/tex] to hit the ground for the second object:

Since both objects are dropped from the same height and air resistance is neglected, the time taken [tex](t_2)[/tex] by the second object to reach the ground is not dependent on its mass and will be the same as the first object.

Thus, the second object will also take 9 seconds to hit the ground.

The acceleration due to gravity (usually denoted as g) is constant at approximately 9.8 m/s² on Earth, which means that in a vacuum, all objects fall at the same rate regardless of their mass.

This principle is often demonstrated in physics classrooms and can be counterintuitive because we see different objects fall at different speeds in everyday life due to air resistance.

An unbalanced force is needed to change its speed or direction. So, option B.

Newton's first law states that, an object will continue its state of rest or uniform motion, unless it is acted upon by an external force.

Here,

The object is said to be moving at a constant speed in one direction. So, a force is required to change the speed or direction of its movement.

Depending upon the mass, an object can be slowed down or accelerated by a force. A force has the power to alter the motion of an object. An object will move differently when subjected to a greater force. In order to experience the same change in motion, a heavier item needs to be subjected to a greater force than a lighter object.

The velocity of an object will alter as a result of unbalanced forces. The object has the ability to alter its speed, direction, or both. An object's velocity changes as a result of unbalanced forces acting on it, which provide a net force.

Hence,

An unbalanced force is needed to change its speed or direction. So, option B.

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Answer:

satphone

Explanation:

plato users

Answer:

Explanation:

sat phone

common knowledge mate

Final answer:

Calculate the time for an object thrown upwards at 16 m/s to reach a height of 7.0 m.The time it takes the object to reach 7.0 m on the way up is approximately 1.63 seconds.

Explanation:

To find the time it takes for the object to reach a height of 7.0 m on the way up:

Identify the known quantities: Initial velocity (u) = 16 m/s, final velocity (v) = 0 m/s (at the top of its flight), acceleration (a) = -9.8 m/s² (acceleration due to gravity).

Use the kinematic equation: v = u + at and solve for time (t).

Calculate the time taken: t = (v - u) / a.

The time it takes the object to reach 7.0 m on the way up is approximately 1.63 seconds.

Final answer:

The speed of sound in the vicinity of a tuning fork with a frequency of 256 Hz and a wavelength of 1.33 m is approximately 341.28 m/s.

Explanation:

The speed of sound can be calculated using the formula:

Speed of sound = frequency x wavelength

Given:

Frequency of the tuning fork = 256 Hz

Wavelength in air = 1.33 m

Substituting the given values, we get:

Speed of sound = 256 Hz x 1.33 m = 341.28 m/s

Therefore, the speed of sound in the vicinity of the fork is approximately 341.28 m/s.

Window glass is a better conductor of thermal energy than a plastic foam cup which acts as a good thermal insulator, slowing down the transfer of heat. The foam cup is better at keeping the heat in, while the window glass allows more heat to pass through.

In the comparison between window glass and a plastic foam cup, the better conductor of thermal energy is window glass. Some materials conduct thermal energy faster than others. Materials like glass and metals are generally good heat conductors, whereas materials such as plastic and rubber are poor heat conductors. Plastic foam like polystyrene foam seen in disposable cups is actually a good thermal insulator, meaning it slows down the transfer of heat. As such, it's great for keeping your coffee hot by reducing heat transfer from the coffee to the environment.

On the other hand, window glass, while not as conductive as metals, still conducts heat at a much faster rate than plastic foam. It is therefore less effective as a thermal insulator, which is why you may feel the cold through a window on a chilly day. Window glass is seen as a major region of heat transfer in buildings, as it allows heat to escape from the interior to the exterior.

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Increasing temperature in lava leads to an increase in its thermal energy, causing more vigorous molecular motion and keeping the lava in a liquid state for a longer period. Cooling of the lava involves energy transfer to its surroundings by radiation and conduction, with notable effects such as steam production when lava encounters colder matter.

As the temperature of lava increases, the thermal energy content of the lava also increases. This increase in thermal energy causes the vibrational motions within the material to become more vigorous. As a result, when we consider the formation of igneous rock, lava that has a higher temperature will remain in a liquid state for a longer period before it cools and freezes into solid rock. The physical processes associated with a large body of cooling lava involve the transfer of energy through radiation and conduction. Specifically, the rate of energy transfer by radiation from the surface of lava into the surroundings can be calculated using the difference in temperature between the lava's surface, its interior, and the surrounding environment.

The scenario detailed here, with lava cooling from 1200°C in the interior to 450°C at the surface, highlights how heat dissipates from the hotter to the colder areas. The process is analogous to when lava flows into cold ocean water, transferring heat so rapidly that steam is produced. Therefore, as lava temperatures increase, the lava will continue to behave as a hotter substance, capable of transferring high amounts of energy to cooler surrounding areas until an equilibrium is reached.

The velocity of the trainee is 29 m/s or 0.42 rev/s

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Centripetal Acceleration of circular motion could be calculated using following formula:

[tex]\large {\boxed {a_s = v^2 / R} }[/tex]

a = centripetal acceleration ( m/s² )

v = velocity ( m/s )

R = radius of circle ( m )

Let us now tackle the problem!

Given:

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

Unknown:

Velocity of Rotation = v = ?

Solution:

[tex]F = ma[/tex]

[tex]F = m\frac{v^2}{R}[/tex]

[tex]7.80w = m\frac{v^2}{R}[/tex]

[tex]7.80mg = m\frac{v^2}{R}[/tex]

[tex]7.80g = \frac{v^2}{R}[/tex]

[tex]7.80 \times 9.8 = \frac{v^2}{11.0}[/tex]

[tex]v^2 = 840.84[/tex]

[tex]v \approx 29 ~m/s[/tex]

[tex]\omega = \frac{v}{R}[/tex]  → in rad/s

[tex]\omega = \frac{v}{2 \pi R}[/tex]  → in rev/s

[tex]\omega = \frac{29}{2 \pi \times 11.0}[/tex]

[tex]\omega \approx 0.42 ~ rev/s[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

The speed of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is [tex]\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex]  and in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is [tex]\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .

Explanation:

The radius of horizontal circle is [tex]11{\text{ m}}[/tex] .and the force is equal to [tex]7.8[/tex]  times the weight of trainee.

Our aim is to obtain the velocity or speed of trainee in both [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  and [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .

The weight of the trainee is calculated as,

[tex]W=mg[/tex]

The force is equal to 7.8 times the weight of trainee and is shown below.

[tex]F=7.8mg[/tex]

The expression for centripetal force is shown below.

[tex]{F_{{\text{centripetal}}}}=\frac{{m{v^2}}}{r}[/tex]                                  ......(1)

The radius of circle is [tex]11{\text{ m}}[/tex] .

The centripetal force is equal to the force exerted by trainee.

So, substitute [tex]7.8mg[/tex]  for [tex]{F_{{\text{centripetal}}}}[/tex]  and [tex]11[/tex]  for [tex]r[/tex]  in equation (1) to obtain the value of velocity in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .

[tex]\begin{aligned}7.8mg&=\frac{{m{v^2}}}{{11}}\\7.8g&=\frac{{{v^2}}}{{11}}\\{v^2}&=85.8g\\\end{aligned}[/tex]

The acceleration due to gravity is [tex]9.8{{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}^{\text{2}}}[/tex] .

Now, the velocity is calculated as,

[tex]\begin{gathered}{v^2}=85.8\left({9.8}\right)\\=840.84\\v=\sqrt{840.84}\\=28.99\\\approx29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{gathered}[/tex]

Therefore, the velocity of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is approximately [tex]29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .

The expression for angular velocity in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is shown below.

[tex]\begin{aligned}\omega&=\frac{v}{2\pi r}\end{aligned}[/tex]         ... (2)

The obtained velocity is [tex]29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] , so substitute [tex]29[/tex]  for [tex]v[/tex]  and [tex]11[/tex]  for [tex]r[/tex]  in equation (2) to obtain the angular velocity.

[tex]\begin{aligned}\omega&=\frac{29}{2\pi(11)}\\&=0.419\\&\approx0.42\text{ rev/s}\end{aligned}[/tex]

Therefore, the angular velocity in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is [tex]0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] .

Thus, the speed of trainee in [tex]{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is [tex]\boxed{29{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex]  and in [tex]{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex]  is [tex]\boxed{0.42{\text{ }}{{{\text{rev}}}\mathord{\left/{\vphantom{{{\text{rev}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .

Learn More:

1. Linear momentum https://brainly.com/question/11947870

2. Motion and velocity https://brainly.com/question/6955558

3. Centripetal Force https://brainly.com/question/7420923

Answer Details:

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords:

Device, astronauts, jet, pilots, rotation, trainee, horizontal, force, weight, fast, m/s, rev/s, tangential, velocity, speed, angular, centripetal.

Option B is correct:)

Sorry im so late!

Answer:

B. The force of friction between the ball and the grass slows the ball to a stop.

Explanation:

When Reyna kicks the ball the force due to contact between her foot and ball accelerate the ball and hence it starts moving on the grass

But after some time ball stops rolling on the grass. So we can conclude here that there must be some external force on the ball due to which it stops rolling on the grass.

As we know that state of motion of an object will change only when some external unbalanced force act on an object.

So here this external unbalanced force is due to grass which resist the motion of ball and it stops.

Here this resistance force is known as friction force of grass on the ball which resist the motion of ball and it stops so correct answer will be

B. The force of friction between the ball and the grass slows the ball to a stop.

The formation of the outer planets are affected by their distance from the sun with having them to maintain the lighter elements that they are composed of such as the hydrogen and helium, having them far away will also make their planet more cooler as the sun is distant from them.

The formation of the outer planets, like Uranus and Neptune, was affected by their distance from the Sun as they likely formed closer to where Jupiter and Saturn are now. The low density of matter in the disk surrounding the Sun outside the orbit of Saturn made it difficult for Uranus and Neptune to form at their present distances. Computer models show that gravitational interactions with neighboring planets could have caused Uranus and Neptune to move to their current locations.

The formation of the outer planets, such as Uranus and Neptune, was affected by their distance from the Sun. According to astronomers, these planets likely did not form at their present distances but closer to where Jupiter and Saturn are now. This is because the density of matter in the disk around the Sun at the time of planet formation was too low outside the orbit of Saturn to build up Uranus and Neptune in a reasonable amount of time. Computer models suggest that Uranus and Neptune could have formed near Jupiter and Saturn and then moved to their current locations through gravitational interactions with their neighboring planets.

The terminal speed of the skydiver falling feet first is 53.2 m/s.

To find the terminal speed of the skydiver, we can use the formula:

vt = sqrt((2 * m * g) / (ρ * A * Cd))

where:

Using the given values, we can plug them into the formula:

vt = sqrt((2 * 72 kg * 9.8 m/s2) / (1.2 kg/m3 * (0.21 m * 0.41 m) * 0.80))

vt = 53.2 m/s

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The terminal speed of a 72 kg skydiver falling feet first is approximately 189.25 m/s. This is calculated using the formula for terminal velocity, with given values for mass, air density, cross-sectional area, and drag coefficient. The cross-sectional area was determined based on the dimensions of the skydiver.

To find the terminal speed ([tex]v_t[/tex]) of the skydiver, we use the formula for terminal velocity:

[tex]v_t[/tex] = √(2mg / ρA[tex]C_d[/tex])

Where:

m = 72 kg (mass of the skydiver)

g = 9.8 m/s² (acceleration due to gravity)

ρ = 1.2 kg/m³ (density of air)

A = 0.041 m² (cross-sectional area in meters squared, which is 0.21 m * 0.41 m)

[tex]C_d[/tex] = 0.80 (drag coefficient)

Putting the values into the formula:

[tex]v_t[/tex] = √((2 * 72 kg * 9.8 m/s²) / (1.2 kg/m³ * 0.041 m² * 0.80))

[tex]v_t[/tex] = √(1411.2 / 0.03936)

[tex]v_t[/tex] = √(35848.78)

[tex]v_t[/tex] ≈ 189.25 m/s

Therefore, the terminal speed of the skydiver falling feet first is approximately 189.25 m/s.

Answer:B

Explanation: Because the weight will be greater with one leg lifted up in the air, it will make the weight more.

Final answer

The wavelength of the light, to the nearest whole number, is 684 nm.

The formula for the diffraction grating is given by d sin( heta) = m extlambda, where d is the distance between adjacent gratings,  heta is the angle of diffraction, m represents the order of the maximum, and  extlambda is the wavelength of the light. For a grating with 200.0 lines per mm, the distance d is the inverse of the number of lines per meter, which is 1/(200,000) meters or 5 x 10^-6 m. The given second-order dark band (m=2) forms at a 20.0°(diffraction angle). Using these values in the diffraction grating equation, solving for  extlambda yields a wavelength of approximately 684 nm. This can be calculated as follows: 5 x 10^-6 m * sin(20.0°) = 2 extlambda, leading to  extlambda approximately equals 684 nm after computation.

Answer:

D. Ultraviolet light

Explanation:

Edge 2021