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kogti [31]
2 years ago
12

Five groups of four vectors are shown below. All magnitudes of individual vectors are equal. Please rank the groups based on the

magnitude of the resultant vectors if the four vectors in each group were added together. Rank the groups of vectors from the greatest resultant magnitude to the smallest resultant magnitude.
Rank the groups of vectors from the greatest resultant magnitude to the smallest resultant magnitude.

Physics
1 answer:
valkas [14]2 years ago
5 0

With the addition of vectors we can find that the correct answer is:

   C)   Q> P > R =  S > T

The addition of vectors must be done taking into account that they have modulus and direction. The analytical method is one of the easiest methods, the method to do it is:

  • Set a Cartesian coordinate system
  • Decompose vectors into their components in a Cartesian system
  • Perform the algebraic sums on each axis
  • Find the resultant vector using the Pythagoras' Theorem to find the modulus and trigonometry to find the direction.

In this exercise indicate that the modulus of all vectors is the same, suppose that the value of the modulus is A.

We fix a Cartesian coordinate system with the horizontal x axis and the vertical y axis, we can see that we do not need to perform any decomposition, so we perform the algebraic sums

Diagram P

x-axis

         x = 2A

y-axis  

         y = 2A

The modulus of the resulting vector can be found with the Pythagorean Theorem

          P = \sqrt{x^2+y^2}

          P = \sqrt{4A^2 +4A^2 }= \sqrt{8}  \  A

          P = 2 √2  A

         

Diagram Q

x-axis

        x = 3A

y-axis  

        y = A

Resulting

       Q = \sqrt{x^2+y^2}

       Q =\sqrt{9A^2 + A^2 }  

       Q = \sqrt{10} \ A

       

Diagram R

x- axis

       x = 0

y-axis

        y = 2 A

Resulting

       R =\sqrt{4A^2 + 0}  

       R = \sqrt{4} \ A

Diagram S

x-axis

       x = 2 A

y-axis

        y = 0

 

Resulting

       S = 2A

Diagram T

x- axis

      x = 0

y-axis  

      y = 0

Resultant T = 0

We order the diagram from highest to lowest

    Q> P> R = S> T

When reviewing the different answers, the correct one is:

   C.  Q> P> R = S> T

Learn more about adding vectors here:

brainly.com/question/14748235

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220V is applied to two different conductors made of the same material. One conductor is twice as long and twice as thick as the
ollegr [7]

Answer:

Explanation:

For calculating resistance of a conductor , the formula is

R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.

For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

R₁ / R₂ = I₁A₂ / I₂A₁

= 2l₂ x A₁ / 4 I₂A₁

= 1 / 2

2R₁ = R₂

Power = V² / R

Ratio of power = (V² / R₁) x (R₂ / V²)

= R₂ / R₁

= 2 .

7 0
3 years ago
What is the objective of playing basagang palayok?the goal what is the goal of playing basang palayok ​
liubo4ka [24]

Answer:

Explanation:

At first glance, it looked like a group of kids-at-heart playing traditional Filipino games under the sun. But up close the games were played by public school teachers staging a unique protest to demand government to prioritize their welfare.

Some 100 members of the Teacher’ Dignity Coalition (TDC) from Metro Manila staged what they called “Palarong Pampista” at the Plaza Miranda in Manila yesterday to demand salary increases this year.

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During the protest, teachers played traditional fiesta games like Palo Sebo and Basagang Palayok “to demonstrate their sacrifices for the country despite the inadequate compensation from the government.”

TDC National Chairperson Benjo Basas said Palo Sebo and Basagang Palayok were symbols of how the government is treating public school teachers.

Palo Sebo, a popular fiesta game in which contestants climb a greased bamboo in order to get the money prize at the top, is the same as the government’s performance-based bonus or PBB, Basas said.

“It promises a cash prize, but teachers and employees need to fight and pull others down in order to be on top,” he explained.

The Basagang Palayok is also another popular fiesta game where the players’ objective is to hit the hanging pot with prizes. Basas said that in this game, “the player is blindfolded, and like the P10,000 salary increase demand, the prize is uncertain and the players must make all the effort and pass the obstacles.”

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“We will not object to any measure that will help public school teachers,” said Luistro. “As long as there is adequate funding, a raise in teachers’ salaries will be welcome,” he said.

However, while DepEd expressed support to the plight of public school teachers for salary increase, Luistro enjoined “certain groups not to take any action that greatly affects the delivery of basic services to our learners.”

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5 0
3 years ago
A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note​
Xelga [282]

Answer:

\huge\boxed{\sf \lambda = 2 m}

Explanation:

<h3><u>Given data:</u></h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3><u>Required:</u></h3>

Wavelength = λ = ?

<h3><u>Formula:</u></h3>

v = fλ

<h3><u>Solution:</u></h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

\rule[225]{225}{2}

8 0
1 year ago
Two radio antennas A and B radiate in phase. Antenna B is a distance of 120 m to the right of antenna A. Consider point Q along
LuckyWell [14K]

Answer:

240 m

120 m

Explanation:

d = Path difference = 120 m

For destructive interference

Path difference

d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m

The longest wavelength is 240 m

For constructive interference

d=\lambda\\\Rightarrow 120\ m=\lambda

The longest wavelength is 120 m

4 0
3 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
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