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vampirchik [111]
3 years ago
13

We emphasize performance over participation well before kids' bodies, minds, and interests mature. And we tend to value the chil

d who can help win games or whose families can afford the rising fees." Take some time to think through your answer.
Physics
1 answer:
Vika [28.1K]3 years ago
4 0

Answer:

Yes.

Explanation:

Yes, we emphasize performance over participation of kids in a competition which causes negative effect on kid's personality. We also prefer those children whose families can afford the fees which is our big failure. We should  emphasize participation over performance and prefer those child who has the ability to win the game instead of those who can afford the fees.

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A student performs experiements on acids and bases. What is not a scientific experiment
seraphim [82]
The answer you are looking for would be C. "She asks her lab partner which base he thinks is hardest to study" 

This is the correct option out of the other choices. 
A. She uses a acid-base indicator to measure the pH of four different solutions 
B. She mixes two solutions and measures their pH before and after 
C. She asks her lab partner which base he thinks is the hardest to study
D. She measures the temperature of a solution before and after adding H2SO4
7 0
3 years ago
Ohm's law relates the current, voltage, and resistance in a circuit. Use Ohm's law to determine what will happen to the remainin
krek1111 [17]

Answer:

   R = ½ R₀

Explanation:

This is an exercise in Ohm's law,

           V = IR

in the initial case

           V₀ = I₀ R₀               (1)

indicates that the voltage remains constant and the current is doubled

             I = 2 I₀

           V₀ = I R

             

we substitute

           V₀ = 2 I₀ R

            R = ½ V₀ / I₀

we replace by equation 1

            R = ½ R₀

8 0
4 years ago
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad
Anna71 [15]

Answer:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

a_{r} = \omega^{2}\cdot R

Where:

\omega - Angular speed, measured in radians per second.

R - Radius of rotation, measured in meters.

The angular speed is first determined:

\omega = \frac{\pi}{30}\cdot \dot n

Where \dot n is the angular speed, measured in revolutions per minute.

If \dot n = 3000\,rpm, the angular speed measured in radians per second is:

\omega = \frac{\pi}{30}\cdot (3000\,rpm)

\omega \approx 314.159\,\frac{rad}{s}

Now, if \omega = 314.159\,\frac{rad}{s} and R = 0.1\,m, the resultant acceleration is then:

a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)

a_{r} = 9869.588\,\frac{m}{s^{2}}

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

a_{r} = 1006.382g \,\frac{m}{s^{2}}

6 0
3 years ago
How much work is done if a 50N weight is lifted to a height of 10m? ​
borishaifa [10]

500J  

Explanation:

Given parameters:

Weight of the body = 50N

Height = 10m

Unknown:

Work done = ?

Solution;

Work done is the force that moves a body through a particular distance in the direction of the force.

In this problem, we can solve the problem by relating work done to the potential energy used in lifting the mass.

  The weight on a body, a force in the presence of gravity

         weight(weight) = mg

  where m = mass of body

              g = acceleration due to gravity

  Work done = Fxd

   Where F = force on the body

               d = distance moved

Potential energy = mgh

    where h is the height

     P.E = work done =  Weight x height = 50 x 10 = 500J  

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

6 0
3 years ago
At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc
amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

F=At^2

Here F = 888.93 N when t = 1.25 s

(c) We can find the value of A first as :

F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2

The value of A is 568.91\ N/s^2.

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

J=\int\limits {F{\cdot} dt}

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns

(b) Impulse is also equal to the change in momentum as :

J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s

Hence, this is the required solution.

5 0
3 years ago
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