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stich3 [128]
3 years ago
14

Explain why you can only change one variable in an experiment

Physics
1 answer:
ohaa [14]3 years ago
5 0
The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to.  But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see. 

Or maybe BOTH of them working together caused it.  You don't know. 

So your experiment is not really much good.  You need to do it again.
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on touching electroscope gets positively charged, so answer is B. conduction

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3 years ago
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
Gennadij [26K]

Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

3 0
3 years ago
Read 2 more answers
After watching a show about submarines, Jamil wants to learn more about the oceans. Which question could be answered through sci
Georgia [21]
The best question that could prompt a scientific investigation is: <u>What substances dissolve in ocean water?</u> 

This way the person experimenting can use several variables and make observations. Upon making observations, the person can be able to gather as much data as he can in order to answer the original question that he asked.
7 0
3 years ago
Read 2 more answers
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0
3 years ago
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