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mylen [45]
3 years ago
11

What investigations allow for the control of variables

Physics
2 answers:
lina2011 [118]3 years ago
7 0
Most as long the hypothesis is a good answer and can be answered 
PolarNik [594]3 years ago
3 0

I believe that the statement is true. Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis

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Which of the following statements describe the transfer of energy?
Lady_Fox [76]

The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.

7 0
3 years ago
In order for gravitational potential energy to be stored, there must be _____.
trasher [3.6K]

Answer: 1

an object positioned at some height in a gravitational field

Explanation:

Gravitational potential energy of an object is the energy stored due to position of the object or position at certain height relative to zero position.

Gravitational potential energy can also be expressed as object position at some height above or below zero position in a gravitational field

I think 1 and 2 make sense. But 1 make more sense than 2

3 0
3 years ago
Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
SHOW WORK
Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\

For Point b:

E=\frac{1}{2} m a^2 w^2\\\\

   =\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J

For Point C:

V_{max}= a w

        = (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}

For point D:

X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm

5 0
3 years ago
Why can't there be a number lower than absolute zero
nexus9112 [7]

Absolute zero is not about numbers.  It's about temperature, and the
motion of molecules in gases. 

You know that the temperature we feel with our skin is the result of the
average speed of all the tiny molecules zipping around or vibrating in
the solid, liquid, or gas.

The faster they're all moving, the warmer the substance feels to us. 
The slower they're all moving, the cooler the substance feels to us.

When molecules slow down to zero and lose all of their kinetic energy,
that temperature is what we call 'absolute zero' ... if they're not moving
at all, then they can't move any slower.

5 0
3 years ago
Read 2 more answers
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