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Nesterboy [21]
3 years ago
5

2. A 0.8 kg tetherball hangs on the end of a cord. It is hit by a child and rises 2.1 m above the ground. a. What is the maximum

gravitational potential energy of the ball?
Physics
1 answer:
skad [1K]3 years ago
3 0

Answer:

E = 16.464 J

Explanation:

Given that,

Mass of tetherball, m = 0.8 kg

It is hit by a child and rises 2.1 m above the ground, h = 21. m

We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :

E = mgh

g is acceleration due to gravity

E = 0.8 kg × 9.8 m/s² × 2.1 m

= 16.464 J

So, the maximum potential energy of the ball is 16.464 J.

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A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

            Force =  Spring constant x Elongation

            F = kx

Here force is weight of mass

           F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 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.

The protest is part of the “Protest de Mayo” launched by TDC earlier this month to call for a P10,000 across-the-board increase in salaries of government teachers and the scrapping of the performance-based bonus of the Aquino government which they described as “deceptive, unfair and divisive.”

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.”

Earlier, Education Secretary Armin Luistro said that the Department of Education (DepEd) “recognizes the right of our teachers to raise their concerns publicly.”

“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.”

TDC, a 30,000-strong group, has set several other unique mass actions until the resumption of classes in June 2 in all public elementary and secondary schools nationwide.

5 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
Describe a situation where you can be traveling at a low speed but have an extremely high velocity
Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0
3 years ago
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Which of the following best describes the velocity of an object
max2010maxim [7]
The definition of volocity is <span>the speed of something in a given direction.</span>
6 0
3 years ago
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