Match the sport/physical activity in column B with the primary physical fitness component needed to perform it in column A . Write the letters of your answer in your activity notebook.

1. Power

A. Patintero

2. Speed

B. Marathon

3. Balance

C. dodgeball / tamaan bata

4. Coordination

D. 100m sprint

5. Flexibility

E. badminton and table tennis

6. Muscular Strength

F. exercise and proper diet


7. Agility

G. hopscotch/piko

8. Cardiorespiratory Endurance

H. Shotput

9. Reaction time

I. Archery


J. Leg Splits and yoga poseso



sagutan po plss

Answers

Answer 1

. Power

Patintero

, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]

2. Speed

The Barkley Marathons

is an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.


Related Questions

At the molecular level, as the kinetic energy increases, what happens to the temperature?

decreases

increases

stays the same

Answers

Answer: increases

Explanation:

Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.

What do alcohol, drugs, and tobacco all have in common?
All have some medicinal value.

All are harmful to the body.

All are depressants.

All are stimulants.

Answers

Answer:

all are harmful to the body

(f) As the pions move away from each other, the ratio of the absolute value of electric potential energy to the final total kinetic energy of the two pions changes. At some point, the potential energy becomes negligible compared to the final total kinetic energy . We can consider that the value of the ratio is about 0.01 when , where is the final total kinetic energy of the two pions. What will the distance, , between the pions be when this criterion is satisfie

Answers

The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

If we consider the potential energy (U) between the pions, then (U) can be expressed as:

[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]

Given that at some instance, the potential energy becomes negligible compared to the final K.E.

As such the conservation of the total energy in the system can be given as:

E = U + K

Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:

[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]

Equating both equations (1) and (2) together, we have:

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]

[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]

[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]

where:

r = distancek = Columb's constantq = charge on a protonm_o = rest mass of each pion in the previous questionc = velocity of light[tex]\mathbf{v_\pi}[/tex] = calculated velocity of proton in the previous question

Replacing their values in the  above equation, the distance (r) between the pions is calculated as:

[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]

distance (r) = 1.45 × 10⁻¹⁶ m

Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m

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Uranus (mass = 8.68 x 1025 kg) and its moon Miranda (mass = 6.59 x 1019 kg) exert a gravitational force of 2.28 x 1019 N on each other. How far apart are they? cs [?] x 10?'m Coefficient (green) Exponent (yellow) Enter​

Answers

Answer:

Explanation:

F = GMm/d²

d = √(GMm/F)

d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)

d = 1.29398e8 = 1.29 x 10^8 m  center to center

Answer:

1.29 x 10^8 m  apart

Explanation:

Works in Acellus!

Just before it strikes the ground, what is the watermelon's kinetic energy?

Answers

Answer:

Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.

Explanation:

This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some

Several common barometers are built using a variety of fluids. For which fluid will the column of fluid in the barometer be the highest

Answers

Answer:

the one in which the fluid has the lowest density

If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power

Answers

Answer correct in this picture………….

The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?

Answers

Answer:

Explanation:

s 0.12 + 1.20(1.80) = 2.28 m from the top.

A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact

Answers

Hi there!

We know that:

I = Δp = m(vf - vi)

Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.

I = 1.1(5 - (-23)) = 30.8 Ns

MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is

1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X​

Answers

Answer:

ma*XsinB

option 2 is correct

A stomp rocket takes 3.1 seconds to reach its maximum height.

- What is its initial velocity? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)

- What is its maximum height? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)

Answers

Answer:

Explanation:

Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.

v = at

v = 9.8(3.1)

v = 30.38

v = 30 m/s

max height can be found knowing the velocity is zero at the top of its flight.

v² = u² + 2as

s = (v² - u²) / 2a

s = (0.00² - 30.38²) / (2(-9.8))

s =  47.089

s = 47 m

A stomp rocket takes 3.1 seconds to reach its maximum height then the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.

What is Velocity?

Velocity is defined as rate of change of position with respect to time.

SI unit of velocity is m/sec. Velocity is a vector quantity.

Given that in the question time taken by rocket to reach maximum height is 3.1 sec. Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.

v = at

v = 9.8(3.1)

v = 30.38

v = 30 m/s

Max height can be found knowing the velocity is zero at the top of its flight.

v² = u² + 2as

s = (v² - u²) / 2a

s = (0.00² - 30.38²) / (2(-9.8))

s =  47.089

s = 47 m

So, the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.

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An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving down at constant speed. What is the tension of the string?

A. Zero
B. Equal to mg
C. Less than mg
D. Greater than mg​

Answers

Answer:

D. Greater than mg​

Explanation:

According to Newton’s second law of motion, the net force equals mass times acceleration. We are going to use a free body diagram (force diagram) to show that the equation of the motion is given by

T – mg = – ma

Thereby,

T = mg – ma

and the answer is: (d)

D. Greater than mg​

_________________________________

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Make one comparison between the moral condition of the world at the time of the Flood with our day. Only One Short explanation.

Answers

The moral condition of the world today appears to be worse than it was in the antediluvian era.

The biblical account of the flood records that the world delved into apostasy in the days of Noah so much so that God regretted the fact that he created man. Some of the evils of the antediluvian world include; sodomy, drunkenness, lewdness and several forms of immorality.

We can see that these vices that led to the destruction of the world due to moral bankruptcy in the antediluvian era is still very much prevalent in our world today. The moral condition of the world today appears to be worse than it was in the antediluvian era.

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The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Answers

We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is

M = 7.30 N.m

From the question we are told

The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A

Generally the equation for the Block is mathematically given as

[tex]\sum Fy=0[/tex]

[tex]981cos21.80 = R_2cos53.6\\\\R_2=1535N[/tex]

the equation for the Wedge is mathematically given as

[tex]\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N[/tex]

the equation for the Screw is mathematically given as

[tex]\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\[/tex]

Therefore

[tex]M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m[/tex]

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A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. How much kinetic energy does the ball have?

Answers

You do 0.5 x 2kg x 4 squared so

0.5x2x16

So the answer is 16

A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).

Answers

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

 

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.​

Answers

Answer:

Explanation:

If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change

EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST

Answers

Explanation:

F = Icurrent×length×Bfieldstrength×sin(angle field to wire)

in our case

Icurrent = 10 A

length = 0.02km = 20 meters

B = 10^-6 T

angle = 30 degrees.

F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =

= 200 × 10^-6 = 2 × 10^‐4 N

A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ride before her cycling computer ran out of battery.

Answers

Answer:

A) 58 km

B) 30 mins

Explanation:

In pic details

graph in pic

Velocity and Acceleration Quick Check

C

D

E

During which of the labeled time segments is the object moving forward but slowing down?

(1 point)

Ο Α

0 С

OD

ОВ

Answers

Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

Which is the most famous book of the philosopher Alexis karpouzos? I think it is the ''Cosmology, philosophy and physics''.

Answers

Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".

Explanation:

Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)

Answers

Its relative speed compared to Earth is 0.921

The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.

Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.

So, v = 2πr/T

Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.

So, v' = 2πr'/T'

v'/v = 2πr'/T' ÷ 2πr/T

v'/v = r'/r × T/T'

From Kepler's law, T² ∝ r³

So, T'²/T² = r'³/r³

(T'/T)² = (r'/r)³

T'/T =  √[(r'/r)]³

T/T' = √[(r'/r)]⁻³

So, substituting this into the equation, we have

v'/v = r'/r × T/T'

v'/v = r'/r × √[(r'/r)]⁻³

v'/v = √[(r'/r)]⁻¹

Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18

So, v'/v = √[(r'/r)]⁻¹

v'/v = √[(1.18)]⁻¹

v'/v = [1.0863]⁻¹

v'/v = 0.921

So, its relative speed compared to Earth is 0.921

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A mass vibrates back and forth from the free end of an ideal spring of spring constant 20 N/m with an amplitude of 0.30 m. What is the kinetic energy of this vibrating mass when it is 0.30 m from its equilibrium position?

Answers

Hi there!

We can begin by using the work-energy theorem in regards to an oscillating spring system.

Total Mechanical Energy = Kinetic Energy + Potential Energy

For a spring:

[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]

A = amplitude (m)

k = Spring constant (N/m)

x = displacement from equilibrium (m)

m = mass (kg)

We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:

ME = KE + PE

ME - PE = KE

Thus:

[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]

Plug in the given values:

[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]

We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.

Thus, the object would have no kinetic energy since KE = 1/2mv².

A crane is lifting a 500 lb car. If the power of the crane 1.82 hp, find the velocity of the car.​

Answers

Answer:

Explanation:

550 ft•lb/s / hp•(1.82 hp) / 500 lb = 2.00 ft/s

How much distance does a car travel with a speed of 2m/s in 15 min?​

Answers

1800m because there 900 seconds in 15 minutes and 2*800=1800 so 1800m

Explanation:

1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2

A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.

The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.

Answers

Answer:

Explanation:

Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.

In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.

That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.

A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?

Answers

Answer:

Explanation:

The work of the brakes will equal the initial kinetic energy of the car

Fd = ½mv²

F = mv²/2d

F = 2457(18²) / (2(62))

F = 6,419.903...

F = 6.4 kN

c. Boat travels north then west

A boat travels 76.0 km due north in 8.0 hours then 56.0 km due west in 5.0 hours.

Determine the direction (as a bearing) of the average velocity (to 1 decimal places) of the boat in the 8 + 5 hour period.


PLEASE HELP!!

Answers

Answer:

Explanation:

θ = arctan(56.0/76.0) = 36.4° West of North

average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s

The oscillation of the 2.0-kg mass on a spring is described by x = 3.0 cos (4.0 t) where x is in centimeters and t is in seconds. What is the force constant k of the spring?

Answers

X is the answer because x can be any thing

The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

What is force?

Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.

Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.

Given:

The mass of the block, m = 2 kg,

The oscillation of spring, x  = 3 cos 4t,

Calculate the omega  by comparing the standard equation given below,

[tex]x = A cos \omega t[/tex]

ω = 4

Calculate the spring constant by the formula given below,

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

4² = k / 2

k = 32 N / m

Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

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A 5 kg box is sitting on a rough wooden surface. The coefficient of static friction between the box and surface is 0.6. If the normal force on the box is 50 N, calculate the force of friction which must be overcome to move the box. Round your answer to the nearest whole number.

Answers

The force of friction needed to overcome to move the box is 29.4N

According to Newton's second law;

[tex]\sum F_x = ma_x\\[/tex]

Taking the sum of force along the plane;

[tex]F_m -F_f = ma\\F_m -F_f = 0\\F_m=F_f = \mu R[/tex]

This shows that the moving force is equal to the frictional force

Given that

[tex]\mu = 0.6\\R = mg = 49N[/tex]

Get the frictional force;

Since

[tex]F_f = \mu R\\F_f = 0.6 \times 49\\F_f = 29.4N[/tex]

Hence the force of friction needed to overcome to move the box is 29.4N

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