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konstantin123 [22]
3 years ago
5

A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to

two groups.students in group one get the tutorial support and those in group two do not.In this example what is dependent variable,independent variable,control groupand experimental group​
Physics
1 answer:
NikAS [45]3 years ago
5 0

Answer:

The dependent variable is academic performance

The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

Explanation:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

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The efficiency of an average gasoline-powered car is about 20%. This means
nydimaria [60]

Answer:

This means C.the car uses 20% of the energy store in the gasoline for motion.

Explanation:

I hope this helps.

4 0
3 years ago
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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
4 years ago
Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.
aliya0001 [1]

Answer:

2.2 x 10-19

Explanation:

Kinetic Energy = 1/2 m v ^2

3 0
3 years ago
How do the tension of the cord and the force of gravity affect a pendulum?
LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But  the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.    

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.    


6 0
3 years ago
Read 2 more answers
A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
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