_____________ is the study of movement in athletes. A. Sports biomechanics B. Posture C. Dynamics D. Anatomy

Answers

Answer 1

Answer:

I believe its A: Sports biomechanics.


Related Questions

Describe the direction in which each object will accelerate. If the object won't accelerate, write "no acceleration." Explain what might be happening to each object in terms of its overall motion.

Answers

Is this photo apart of this question? Or you need answer for just the sitting part

if A and B are non zero vectors, is it possible for vector A×vector B and vector A.vector B both to be zero? Justify your answer​

Answers

Answer:

not able to understand the question

how far will an object fall in 3 seconds

Answers

Answer:

144.7832 ft

Explanation:

heres the formula i found to solve

v = v₀ + gt

where:

v₀ is the initial velocity (measured in m/s or ft/s);

t stands for the fall time (measured in seconds); and

g is the free fall acceleration (expressed in m/s² or ft/s²).

Without the effect of air resistance, each object in free fall would keep accelerating by 9.80665 m/s (approximately equal to 32.17405 ft/s) every second. In reality, though, a falling object's velocity is constrained by a value called the terminal velocity.

im sorry i couldnt provide more information, but there was not much info provided in the question itself

The Earth currently has an orbital period of 1 year and a semi-major
axis length of 1 AU. If the Earth's orbit changed and the semi-major
axis length increased to 1.1 AU, how would the Earth's orbital period change?

A: It would increase by 0.15 years

B: It would increase by 0.032 years

C: It would decrease by 0.15 years

D: It would decrease by 0.032 years

Answers

The Earth's orbital period change is A: It would increase by 0.15 years

Using Kepler's third law which states that the square of the orbital period of the planet is directly proportional to the cube of its distance from the sun.

So, T² ∝ R³

T'²/T² = R'³/R³ where

T = orbital period at R = 1 yearR = initial axis length = 1 AUT' = orbital period at R'R' = final axis length = 1.1 AU.

So, making T' subject of the formula, we have

T' = [√(R'/R)³]T

T' = [√(1.1 AU/1 AU)³] × 1 year

T' = [√(1.1)³] × 1 year

T' = √1.331 × 1 year

T' = 1.15 × 1 year

T' = 1.15 years.

So, the change in the Earth's orbital period ΔT = T' - T

= 1.15 years - 1 year

= 0.15 years

Since this is positive, the orbital period increases by 0.15 years.

So, the Earth's orbital period change is A: It would increase by 0.15 years

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A racing car has a mass of 750 kg. It undergoes an acceleration of 4.00 m/s2. What is the net force acting on the car?

Answers

Answer:

3000 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 4 = 3000

We have the final answer as

3000 N

Hope this helps you

1)A chocolate bar measures 10 cm long by 2 cm wide and is 2 cm thick
a)Calculate the volume of one bar.

b)How many bars each 2 cm long, 2 cm wide and 2cm thick have the same total volume?

c)A pendulum makes 10 complete oscillations in 8 seconds. Calculate the time period of the pendulum​

Answers

Answer:

40cm³

Explanation:

I only have an answer to 1a

Volume = Length X breadth X height

where;

length = 10cm

breadth = 2cm and;

width = 2cm

*Find*

V = ?

V = L X B X H

V = 10 X 2 X 2

= 40cm³✓



A roller coaster goes from 2.00 m/s [forward] to 10.0 m/s [forward) in 4.50 s. What is its acceleration?

Answers

The answer would be 1.77778:

We must use the kinematic equation v = v0 + at, then fill in the elements given and solve which equals 1.77778

Which is a possible food chain in the ocean from start to finish?
a.phytoplankton, zooplankton, small fish, large fish, large shark
b.phytoplankton, large shark, zooplankton, large fish, small fish
c.large shark, large fish, small fish, zooplankton, phytoplankton
d.zooplankton, phytoplankton, large fish, small fish, large shark

Answers

Answer:

a.phytoplankton, zooplankton, small fish, large fish, large shark

Explanation:

The bottom level of the ocean's food chain is made up of one-celled organisms called phytoplankton. These tiny organisms are microscopic. They are so small they cannot be seen without a microscope. Billions of phytoplankton live in the upper part of the ocean. They take in the sun's light.

Phytoplankton, or plant plankton, is the first link in the food chain in the ocean. They require sunlight for photosynthesis, which they must perform. They therefore reside higher up in the shallow sunlight. Furthermore, they transform themselves into an ocean animal that can eat the “food” from the sun. Thus, option A is correct.

What food chain in the ocean from start to finish?

One-celled organisms known as phytoplankton make up the base of the ocean's food chain. These microscopic organisms are quite small. Without a microscope, they are invisible due to their small size. In the upper ocean, there are billions of phytoplanktons. They absorb the light from the sun.

Aquatic food webs are built on phytoplankton and algae. They are consumed by zooplankton, tiny fish, and crustaceans, which are the main consumers. Fish, tiny sharks, corals, and baleen whales all devour the primary consumers after them.

Therefore, phytoplankton, zooplankton, small fish, large fish, large shark  is a possible food chain in the ocean from start to finish.

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#SPJ2

A wave has wavelength of 10 m and a speed of 340 m/s. What is the frequency of the wave?
I need the Formula,Known,Substitute & Solve Answer with Units

Answers

Answer:

This is the answer that I got.

Explanation:

Hope it is right.

A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?

Answers

The final speed of the combined masses is 0.21 m/s

Applying the law of conservation of momentum:

Total momentum before collision = Total momentum after collision.

⇒ Formula:

MU+mu = V(M+m).................. Equation 1

⇒ Where:

M = mass of the first bodym = mass of the second bodyU = Initial speed of the first bodyu = Initial speed of the second bodyV = common final speed.

From the question,

⇒ Given:

M = 0.24 kgU = 0.60 m/sm = 0.26 kgu = -0.20 m/s (traveling in opposite direction)

⇒ Substitute these values into equation 1

0.24(0.6)+0.26(-0.20) = V(0.24+0.2)

⇒ Solve for V

0.144-0.052 = 0.44V0.44V = 0.092V = 0.092/0.44V = 0.209V ≈ 0.21 m/s

Hence the final speed of the combined masses is 0.21 m/s

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A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to the cup or surroundings. What is the final temperature of the mixture? answer in celsius​

Answers

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

What chemical reactions support all life on earth?

Answers

Answer:

I’d say both anabolic and catabolic chemical reactions make life on Earth possible. Anabolic is the building up of molecules from atoms or small molecules and catabolic is the chemical reactions of breaking them down.

Explanation:

Learning Task 1: Define or describe the following words relat are not familiar with the word, you may ask assistance from y your answer in your notebook. 1. table 2. tennis C 3. ping-pong 4. net 5. ball 6. racket​

Answers

Answer:

1.) TABLE = The table is 2.74 m (9.0 ft) long, 1.525 m (5.0 ft) wide, and 76 cm (2.5 ft) high with any continuous material so long as the table yields a uniform bounce of about 23 cm (9.1 in) when a standard ball is dropped onto it from a height of 30 cm (11.8 in), or about 77%.

2.) TENNIS = Tennis is a racket sport that can be played individually against a single opponent (singles) or between two teams of two players each (doubles). Each player uses a tennis racket that is strung with cord to strike a hollow rubber ball covered with felt over or around a net and into the opponent's court.

3.) PING-PONG = Table tennis, also known as ping-pong and whiff-whaff, is a sport in which two or four players hit a lightweight ball, also known as the ping-pong ball, back and forth across a table using small rackets. ... Spinning the ball alters its trajectory and limits an opponent's options, giving the hitter a great advantage.

4.) NET = This is stretched across the centter of the table by a cord attached to a post at either end. It measures 6ft long and the ball must pass over it for a rally to continue.

5.) BALL = The ball, which is spherical and hollow, was once made of white celluloid. Since 1969 a plastic similar to celluloid has been used. The ball, which may be coloured white, yellow, or orange, weighs about 0.09 ounce (2.7 grams) and has a diameter of about 1.6 inches (4 cm).

6.) RACKET = A table tennis racket is made up of two distinct parts - a wooden blade which incorporates the handle together with table tennis rubbers affixed to each side of the blade using water-based glue.

Explanation:

All my answers are

about the tools of the

game of tennis

What is the radius of a circular space station that spins with a linear speed of 90.0 m/s I’m order for its walls to supply a centripetal acceleration equal to the acceleration provided by gravity on the surface of the earth?

Answers

Answer:

a = v^2 / R      centripetal acceleration

R = v^2 / a     a at the surface of earth is 9.8 m/s^2

R = (90 m/s)^2 / 9.8 m/s^2 = 827 m

which constellation is in contrast with ursa major

Answers

Answer: Ursa Major (/ˈɜːrsə ˈmeɪdʒər/; also known as the Great Bear) is a constellation in the northern sky, whose associated mythology likely dates back into prehistory. Its Latin name means "greater (or larger) she-bear," referring to and contrasting it with nearby Ursa Minor, the lesser bear.

Bordering constellations: Draco; Cameloparda...

Brightest star: ε UMa (Alioth) (1.76m)

Meteor showers: Alpha Ursa Majorids; Leonid...

Symbolism: the Great Bear

A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?

Answers

The  final speed of the combined masses is 0.186m/s

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

Mathematically;

m1u1 - m2u2 = (m1+m1)v

v is the final speed of the combined masses.

Substituting the given parameters;

0.24(0.6) - 0.26(0.2) = (0.24+0.26)v

0.144 - 0.052 = 0.5v

0.092 = 0.5v

v = 0.092/0.5

v = 0.184m/s

Hence the final speed of the combined masses is 0.186m/s

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Please answer the following question!
What is the momentum of a 750-kg Volkswagen Beetle when at rest?
Please give the value and momentum!

Answers

Answer:

0

Explanation:

the momentum will always be 0 when it is at rest because the object isnt moving!

Hope this helped!

What do you feel when you receive your homework? ​

Answers

Very bad, it’s extremely draining I dread it a lot maybe it’s just cause I have a lot

The feeling can either be positive or negative.

How one feels depends on number of factors

A positive feeling can occur when one performs very well in the given home. Also, a positive feeling can come from a high level of satisfaction in the homework. When you complete your homework on time using the recommended steps, you will be sure to do well on the homework.

In other hand, a negative feeling may result from poor performance in the homework. In ability to complete the homework or missing some steps in the homework can increase your level of trepidation even before seeing your score.

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What is number 10 for this?

Answers

Answer:

The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Given data in the question;

Speed of wave;

Frequency of wave;

wavelength;

To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.

Where  is wavelength, f is frequency and c is the speed.

We substitute our given values into the equation

Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Explanation:


1. Which of the following types of energy is
potential energy? More than one answer is
possible,
A kinetic energy
B.thermal energy
C.sound energy
D.gravitational energy

Answers

Answer:

Potential energy is stored energy and the energy of position––gravitational energy. There are several forms of potential energy. Electrical Energy is the movement of electrical charges. Everything is made of tiny particles called atoms.

“Applications of thermal expansion in solids, liquids and gases.”
(Two for each states of matter)​

Answers

Thermal expansion is the tendency of matter to change its shape dimensions, density in response to a change in temperature.

The following are applications of thermal expansion in various matter.SolidsThe principle of thermal expansion is used in thermostat to regulate the temperature of heating device e.g Electric pressing Iron.Rivets are used to hold steel plates together very tightly.

 

  2. Gases

Hot-air balloons are an application of thermal expansionGases turbine

    3. Liquids

Engine CoolantsVolume of thermometer

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Which image shows an example of the electromagnetic force in action?

Answers

Answer:

Where are the images?

Explanation:

I can't help if there is no image(s) to this question.

anything that has mass and occupies space is called

Answers

Anything that has mass and occupies space is called matter.Matter is of three types:SolidLiquidGas

Answer:

Matter

Hope you could get an idea from here.

Doubt clarification - use comment section.

The instant before a batter hits a 0.14-kilogram baseball, the velocity of the ball is 45 meters per second west. The instant after the batter hits the ball, the ball's velocity is 35 meters per second east. The bat and ball are in contact for 1.0 x 10-2 second.
1. find the magnitude and direction of the average acceleration of the baseball while it is in contact with the bat.
2. calculate the magnitude of the average force the Bat exerts on the ball while they are in contact. ​

Answers

(1) The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is 8,000 m/s² east.

(2) The magnitude of the average force the Bat exerts on the ball while they are in contact is 1,120 N.

The given parameters:

Mass of the baseball, m = 0.14 kgInitial velocity of the baseball, u = 45 m/s west (negative direction)Final velocity of the baseball, v = 35 m/s east (positive direction)Time of contact, t = 0.01 s

The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is calculated as follows:

[tex]a = \frac{v- u}{t} \\\\a = \frac{35 - (-45)}{0.01} \\\\a = \frac{80}{0.01} \\\\a = 8,000 \ m/s^2[/tex]

The magnitude of the average force the Bat exerts on the ball while they are in contact is calculated as follows;

[tex]F = ma\\\\F = 0.14 \times 8,000\\\\F = 1,120 \ N[/tex]

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Here's a question from ~ [ AIEEE 2002 ]

The minimum velocity ( in m/s ) with which a car driver must traverse a flat curve of radius 150 m and Coefficient of friction 0.6 to avoid skidding is ~

[ I'm looking for Proper Information, and please don't get it from any Website ]

Thanks for Answering !​

Answers

r=150mcoefficient of friction=[tex]\mu[/tex]=0.6

As car is avoid skidding

[tex]\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg[/tex]

Cancel m

[tex]\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g[/tex]

[tex]\\ \sf\hookrightarrow v^2=\mu rg[/tex]

[tex]\\ \sf\hookrightarrow v^2=0.6(10)(150)[/tex]

[tex]\\ \sf\hookrightarrow v^2=60(150)[/tex]

[tex]\\ \sf\hookrightarrow v^2=900[/tex]

[tex]\\ \sf\hookrightarrow v=30ms^{-1}[/tex]

Done

The minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is 29.7 m/s.

The given parameters:

Radius of the curve, r = 150 mCoefficient of friction, μ = 0.6

The minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is calculated as follows;

[tex]\frac{mv^2}{r} = \mu mg\\\\v^2 = \frac{\mu mgr}{m} \\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr} \\\\v = \sqrt{0.6 \times 9.8 \times 150} \\\\v = 29.7 \ m/s[/tex]

Thus, the minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is 29.7 m/s.

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a water balloon is thrown a target at 18 m/s. if the water balloon has a mass of 0.4 kg, what is the momentum?

Answers

The momentum of the water balloon at the given speed is 7.2 kgm/s.

The given parameters:

Speed of the water balloon, v = 18 m/sMass of the balloon, m = 0.4 kg

The momentum of the water balloon is calculated as follows;

P = mv

Substitute the given values of mass and velocity as follows;

P = 0.4 x 18

P = 7.2 kg m/s.

Thus, the momentum of the water balloon at the given speed is 7.2 kgm/s.

Learn more about momentum here: https://brainly.com/question/7538238

can you fit all the planets between earth and moon

Answers

Answer: No, planets of our solar system, with or without Pluto, cannot fit within the mean lunar distance. An additional 3,500 km is needed to squeeze in Neptune

Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)​

Answers

Happy Holidays!

We can use the following equation to solve for the gravitational force:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Fg = force due to gravity (N)

G = Gravitational constant

m1,m2 = masses of the objects (kg)

r = distance between the objects (m)

Plug in the given values into the equation:

[tex]F_g = (6.67*10^{-11})\frac{(70)(60)}{(2)^2}} = \boxed{7.0 * 10^{-8}N}[/tex]

a truck in a traffic circle travels in a circular path at constant speed v0 while passing through region x of the circle, as shown in figure 1 above. a diagram representing the forces exerted on the truck in region x is shown in figure 2. at a later time a car with less mass than the truck passes through region x at the same speed and the same distance from the center of the traffic circle as the truck. the coefficient of friction between the car’s tires and the ground is the same as that for the truck’s tires and the ground. which of the following diagrams could represent the forces exerted on the car in region x compared to the truck in region x ? assume that the length of each arrow is proportional to the magnitude of the force represented by the arrow.

Answers

The Diagram that represents the forces exerted on the car in region x when compared to the truck in region X is :  Diagram C

Given that the mass of the car is less than the mass of the truck the force due to gravity will be greater on the truck than it could be on the car

i.e. | Fg |car  <  | Fg | truck

also : | Fₙ | car  <  | Fₙ | truck

Since the gravitational force on the car is less than the gravitational force on the truck

| Ff | car < | Ff | truck

where : Ff = frictional force

             Fg = force due to gravity

             Fn  normal force

Therefore from the diagrams attached below the correct diagram that represents the forces exerted on the car in region x when compared to the truck in region X is :  Diagram C.

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Attached below is the missing data related to your question

4 What type of circuit is described by each of the following statements?
Answer series or parallel
a All components are connected end-to-end.
b. The current in the circuit divides so that some flows through one component
and the rest through another component
Two lamps are connected side by side so that each lights brightly
d The current has the same valur everywhere in the circuit
C

Answers

b. the current in the circuit divides so that some flows through one component
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