A. praised
B. scolded
C. rewarded
14. Reicon's parents admonished him for breaking their antique collection,
D. silenced
15. I am an advocate of anti-corruption movement in the country.
D. supporter
C. enemy
DirectioRead the statements below. Using context clues, identify the meaning
A. opponent
B. primer
highlighted words.
16. The hotel abolishes its rice-all-you-can policy to avoid wastage of food.
b. removes
d. proposes
A. continues
17. The country has a very tropical climate where it only has dry and wet
seasons similar to other countries in Southeast Asia
a. humid
b. cold
d. normal
-
c. comfortable
d. collected
a. new
c. outdated
18. The obsolete books and other references in the library resulted to the
removal in the shelves and inventory.
b. old
19. Beverages, such as juices, milk, beers and sodas, have increased in
,
due to low supply and importation issues.
a. drinks b. food
d. pastas
20. The luxurious life of my cousin in an exclusive subdivision in Ta
makes me inspired to dream and work hard for me to escape this unp
d.
a. ordinary b. fantastic
c. uncomfortable
c. bread








Help help me plis po summative test ko po ito plis help me

A. PraisedB. ScoldedC. Rewarded14. Reicon's Parents Admonished Him For Breaking Their Antique Collection,D.

Answers

Answer 1

Answer:

14. scolded

15. supporter

16. i am not sure but I think it is removes

17. comfortable

18. outdated

19. drink

20. it can be fantastic or wealthy

Answer 2

Answer:

14. Reicon's parents admonished him for breaking their antique collection,

scolded

15. I am an advocate of anti-corruption movement in the country.

supporter

16. The hotel abolishes its rice-all-you-can policy to avoid wastage of food.

it is removes

17. The country has a very tropical climate where it only has dry and wet

seasons similar to other countries in Southeast Asia

comfortable

18. The obsolete books and other references in the library resulted to the

removal in the shelves and inventory.

outdated

19. Beverages, such as juices, milk, beers and sodas, have increased in,due to low supply and importation issues.

drinks

20. The luxurious life of my cousin in an exclusive subdivision in Ta makes me inspired to dream and work hard for me to escape this unp.

fantastic

Related Questions

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

. 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.

What is the direction of the torque produced on the crankset by the 2-kg mass attached to the pedal bar

Answers

A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.

The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.

Counterclockwise is the positive rotation direction and clockwise is the negative direction.

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How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves away from a stationary sound source

Answers

Answer:

If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.

what memory are you using to remember who the president of the united states is

Answers

Answer:

The First 8 Presidents

For this exercise, we're going to use a silly story made of silly sentences. The letters that represent the last names of these presidents are W, A, J, M, M, A, J, V. One silly sentence to help you remember this sequence is: Wilma and John made merry and just vanished

working memory.

sensory memory.

short-term memory.

long-term memory.

I need help been struggling on this question

Answers

Answer:

440 m

Explanation:

S=(u+v) t / 2

S = (11+33) × 20/2

S= 44× 20/2

S=440 m

This is two or more elements chemically combined in a fixed ratio.


Example: water, carbon dioxide, sodium chloride

Answers

A compound is a substance that contains two or more elements chemically combined in a fixed proportion. The elements carbon and hydrogen combine to form many different compounds.

Acceleration of a Car A car traveling along a straight road at accelerated to a speed of over a distance of ft. What was the acceleration of the car, assuming that it was constant

Answers

Answer:

how many feet?

Explanation:

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

Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.

Answers

Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.

Answers

Answer:

in general context yes it is closest to 1.04

Explanation:

theres no right or wrong way to scientifically prove this though.

Overall in scale its closest to 1.04 hope that helped

Convert 6 picoseconds into seconds.

Answers

Answer:

6e-12

Explanation:

divide the time value by 1e+12

how do all organisms begin life

Answers

Answer:

All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell

may change little during their lives, but they do grow

Explanation:

brainlest me please

14 The radius of gyration of a body about an axis &ta
distance 6 cm from its centre of mass is 10 cm.
Then, its radius of gyration about a parallel axis
through its centre of mass will be
(a) 80 cm (b) 8 cm (c) 0.8 cm (d) 0.08 cm

Answers

Correct option is B 8 cm.

Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.

Parallel axes theorem gives:

[tex] {mr}^{2} = m( {k}^{2} + {a}^{2} ) \\ ⇒ {r}^{2} = {k}^{2} + a {}^{2} [/tex][tex]⇒k = \sqrt{ {10}^{2} - {6}^{2} } = 8cm.[/tex]

Thus, option B is the correct answer.

An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?

Answers

Hi there!

We can use work and energy to solve this problem.

We know that:

Ei = Ef

Ei = Potential energy = mgh

Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²

The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:

I (hollow cylinder) = mr²

I (disk) = 1/2mr²

Calculate the moment of inertias of each.

Since the mass on the base is one-fourth of its side:

x = mass of side

x + x/4 = 15

4x + x = 60

5x = 60

x = 12 kg

end mass = 3 kg

Solve for each moment of inertia:

Side: (12)(0.4²) = 1.92 Kgm²

Bottom: 1/2(3)(0.4²) = 0.24 Kgm²

Side + bottom = 2.16 Kgm²

We can now solve:

mgh = 1/2mv² + 1/2(2.16)v²/r²

(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²

4851 = 14.25v²

v = 18.45 m/s

PLEASE HELP!
A 9kg particle is initially at rest at x=0. It is subject to a single force Fx (N) which varies with x (m) as shown in the
diagram
F
2
1
0
ББ by
x
- 1
1
-2
The kinetic energy of the particle when it is at x = 3 m is:

Answers

Hi there!

With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.

We know that:

W = ΔKE

Calculate the work by finding the area underneath the force curve from

x = 0 to 3 m:

We can use a trapezoid:

A = 1/2(3 + 2)(3) = 7.5 J

This is the amount of work done, and since the object starts from rest:

7.5J = KEf - KEi (0 J)

7.5J = KEf = 7.5 J

Find the time it takes for an object dropped from a building and reaches a final velocity of 20 m/s downward?

I need the formula

Answers

Answer:

Explanation:

v = at

t = v/a

t = 20 m/s / 9.8 m/s²

t = 2.0408163...

t = 2.0 s

Accelerations are produced by

A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces

Answers

The correct answer is D.
Acceleration is produced by a net force on an object.

plz help me on this question thank you

Answers

Answer:

D

Explanation:

A ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s​

Answers

Answer:b

Explanation:

-29.4 m/s

The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.

What is velocity ?

Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.

Given that, height of the building = 80 m

the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².

Then after 3 seconds, the velocity of the ball is calculated as follows:

velocity = acceleration × time

v = 9.8 m/s²  × 3 s = -29.4 m/s

If the ball reaches the ground within the time of 3 s, then, the velocity is:

v = 80 m/3s = 26.6 m/s.

Find more on velocity :

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#SPJ6

100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.
1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?
2. At point C, the rock's potential energy will be
3. The rock's kinetic energy at point A is
4. At point B, some of the rock's potential energy will be changed to Kinetic energy
5. What is the mass of the rock?
6. What is the rock's velocity just before it hits the ground?
The rock to the right is sitting at the top of a ramp. I wonder how much work it required to get that rock up there.

Answers

Answer:

lol

Explanation:

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​

_________________________________

(hopet his helps can I pls have brainlist  (crown)☺️)

Children in a tree house lift a small dog in a basket 3.85 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket

Answers

Answer:

Explanation:

The work will equal the increase in potential energy.

PE = mgh

m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg

An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]

Answers

a.

The object's speed at 2.20 m below balcony level is 8.74 m/s

Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.

Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and

So, E = E'

U + K + f = U' + K' + f'

where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).

So,

U + K + f = U' + K' + f'

mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh

mgh = mgh' + 1/2mv² + 0.10mgh

Dividing through by m, we have

gh = gh' + 1/2v² + 0.10gh

So, gh -  0.10gh = gh' + 1/2v²

0.90gh = gh' + 1/2v²

1/2v² = 0.90gh - gh'

1/2v² = g(0.90h - h')

v² = 2g(0.90h - h')

Taking square-root of both sides, we have

v = √[2g(0.90h - h')]

where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and  g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

v = √[2g(0.90h - h')]

v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]

v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]

v = √[2 × 9.8 m/s²(3.901 m)]

v = √[76.4596 m²/s²]

v = 8.74 m/s

So, the object's speed at 2.20 m below balcony level is 8.74 m/s

b.

Yes it does matter when we apply 10% loss before V calculations

We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.

So, yes it does matter when we apply 10% loss before V calculations

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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

Check Pic please, need help immediately ​

Answers

It’s d


I did this already


3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1

Answers

Answer:

a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over

Answers

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection​? why

Answers

Answer:

Explanation:

The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity.   The deflection is also inversely proportional to its mass.   So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.

Nikolai is using a hand-operated grain mill to grind wheatberries into flour. The mill is operated by spinning a fly-wheel with radiusR= 23 cm, which has a handle attachedto the outer edge. After grinding for a few minutes at a con-stant angular speedωi, he lets go of the handle and allows themechanism to come to rest as it undergoes constant angularacceleration. This happens over the course oft= 0.50 s, andthe flywheel undergoes a quarter of a rotation during this time.What is the linear tangential accelerationaof the handle as itcomes to rest? For the limits check, investigate what happenstoaas the time required to stop the flywheel becomes small(t→0).

Answers

Answer:

Explanation:

α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²

ωf² = ωi² + 2αθ

θ = (ωf² - ωi²) / 2α

2π/4 = (0² - ωi²) / (2(-2ωi))

2π/4 = ωi / 4

ωi = 2π rad/s

α = -2(2π) = -4π rad/s²

a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²

as the time to stop approaches zero, acceleration goes toward infinity.

PLEASE HELP ON THIS QUESTION ​

Answers

[tex]r = 1.29×10^8\:\text{m}[/tex]

Explanation:

According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is

[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]

where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get

[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]

Taking the square root of the equation above, we get

[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]

Plugging in the values, we get

[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]

[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]

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