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3 years ago

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A characteristic in an experiment, such as how much food, how much space, how hot or how cold, is termed a ____________. (spelli

ng counts, so be careful!)

2 answers:

KATRIN_1 [288]3 years ago

6 0

I think the term is a variable?

yuradex [85]3 years ago

3 0

Variables are what we change in an experiment to test their effects on whatever it is we are measuring. Hope that helps!

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Intro to Springs and Hooke’s law

wel

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.
There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the elasticity is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.

6 0

3 years ago

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

7 0

1 year ago

If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p

Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0

3 years ago

Please help me with this one

katen-ka-za [31]

Both of the pictures show how one picture is I believe in fall and the other one is in the forest and maybe in spring or summer. Also, the two pictures show 2 different animals. And maybe hunting for 2 different foods for them to survive.

* Hopefully this helps:) Mark me the brainliest:)

5 0

3 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago