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ivanzaharov [21]
3 years ago
5

A kind of variable that a researcher purposely changes in an investigation is constant. controlled. dependent. independent.

Physics
1 answer:
AysviL [449]3 years ago
3 0
I think the correct answer would be the last option. A kind of variable that a researcher purposely changes in an investigation is the independent variable. It is the value that does not depend to any change in any variable involved. Instead, it is the one that affects the other variable called the dependent variable. The dependent variable is the variable that is being tested or studied. Also, along with these two there are variables that are held equal all through out the research. These are called the constants. These constants are factors that might affect the dependent variable that is why they are held constant to prevent more errors.
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gregori [183]
There are 3 bases before you reach home plate.
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3 years ago
A drum rotates around its central axis at an angular velocity of 19.8 rad/s. If the drum then slows at a constant rate of 3.02 r
Debora [2.8K]

a) Time taken =6.39 s

b) Angle of rotation while coming to rest = 188.17 rad

<u>Explanation:</u>

Initial angular velocity of the drum \omega_{0} = 19.8 rad/s^{2}

Angular acceleration of the drum \alpha = -3.02 rad/s^{2}

a.) We know that

\omega =\omega_{0} + \alpha  t

0 =  19.8 - 3.02 t

t = \frac{19.8}{3.02}

t = 6.39s

b.) we know that

\theta = \omega_{0} t + \frac{1}{2} \alpha   t^{2}

\theta = 19.8 \times 6.39 + \frac{1}{2} \times 3.02 \times 6.39^{2}

\theta =188.17 rad

3 0
4 years ago
A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340
Fantom [35]

Answer:

The value is     P_G  =  2.925 *10^{5} \  Pa

Explanation:

From the question we are told that

   The  volume of the bottle is  v  =  2 \  L  =  2 *  10^{-3} \  m^3

   The gauge pressure of the air is P_g  =  340 \  kPa  =  340 *10 ^{3}  \  Pa

   

Generally the volume of air before the bottle is turned upside down is  

      V_a  =  \frac{3}{4}  * V

       V_a  =  \frac{3}{4}  *  2 *10^{-3}

       V_a  =  0.0015 \  m^3 }

Generally the volume air when the bottle is turned upside-down is

      V_u  =  \frac{5}{6}  *  2 *10^{-3}

       V_u  =  0.00167 \  m^3

From the the mathematical relation of adiabatic process we have that

     P_g *  V_a^r  =  P_G *  V_u^r

Here r is a constant with  a value  r =  1.4

So

      340 *10 ^{3}  *  0.0015^{1.4}  =  P_G *  0.00167^{1.4}

        P_G  =  2.925 *10^{5} \  Pa

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