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Lubov Fominskaja [6]
3 years ago
12

An experiment is designed to test the relationship between the initial height of a basketball before it is dropped to the height

of its rebound bounce. The height of the rebound bounce is measured using a scale positioned behind the ball.
In the above experiment, which condition would not be controlled?
Physics
2 answers:
iren [92.7K]3 years ago
6 0
In general, when coming up with the experiment, it's fascinating to control all factor which will have an effect on the end result of the experiment. To observe, this is often not simply gettable.
I imagine it'd be hard to manage the temperature during this experiment. although, the temperature will have an effect on the ball's bounce. 
Keep in mind that even if you do not management one thing it still is often measured then you'll be able to use that information to<span> correct your predictions.
Explanation:

</span>Now an uncontrolled variable is something like a force of nature, size of one thing or it's physically that cannot be controlled or <span>modified.

</span>An uncontrolled variable, or intermediator<span> variable, </span>is that the<span> variable in </span>an associate<span> experiment that has the potential to negatively impact </span>the link<span> between the </span>independent<span> and dependent variables.</span>


Zina [86]3 years ago
4 0
In general, when designing the experiment, it is desirable to control all thing that can affect the outcome of the experiment. In practice, this is not easily obtainable.
I imagine it would be hard to control the temperature in this experiment. Even though, temperature does affect the ball's bounce. 
Keep in mind that even if you don't control something it still can be measured and then you can use that information to correct your predictions.
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Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhan
AVprozaik [17]

Answer:

84%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

y_0=1\ m

x = 10 m

t = Time taken

v_0 = 3.5 m/s (assumed, as it is not given)

A_0=\pi 1.5^2

We have the equation

y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s

x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s

From continuity equation we have

Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2

Fraction is given by

f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%

The fraction is 84%

8 0
3 years ago
Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

x = -1.12 m

5 0
3 years ago
Which is one property of coal that makes it ideal for use in a power plant?
photoshop1234 [79]

Answer:

It burns easily & It produces a lot of energy

Explanation:

3 0
4 years ago
An object is removed from a room where the temperature is 69 degrees and is taken outside, where the air temperature is 30 degre
Yuliya22 [10]

Answer:

The temperature of the object at any time t, T(t) is given as

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

Explanation:

Let T be the temperature of the object at any time

T∞ be the temperature outside = 30°

T₀ be the initial temperature of the object in the room = 69°

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the object = Rate of Heat gain by the outside air

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 30) = (69 - 30)e⁻ᵏᵗ

(T - 30) = 39 e⁻ᵏᵗ

At 1 minute, T = 52°

52 - 30 = 39 e⁻ᵏᵗ

22/39 = e⁻ᵏᵗ

- kt = In (22/39) = In (0.564)

- k(1) = - 0.5725

k = 0.5725 /min

(T - T∞) = (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

T = T∞ + (T₀ - T∞)e⁻⁰•⁵⁷²⁵ᵗ

4 0
3 years ago
Relating the circumference to the force applied on the balloon, make a conclusion on the relationship of force to the rate of th
natka813 [3]

At a constant force, the mass of the balloon is inversely proportional to the rate of change motion of the balloon.

The force applied to an object can be determined by applying Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

where;

  • <em>m is the mass of the balloon</em>
  • <em>a is the change in velocity per time</em>

F = m\frac{\Delta v}{t} \\\\\frac{F}{\frac{\Delta v}{t}} = m

The mass of an object is inversely proportional to the rate of change motion of the object.

Thus, we can conclude that at constant force, the mass of the balloon is inversely proportional to the rate of change motion of the balloon.

Learn more here:brainly.com/question/15321240

3 0
3 years ago
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