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antoniya [11.8K]
4 years ago
12

In a controlled experiment do none of the variables change?

Physics
2 answers:
tangare [24]4 years ago
8 0

Answer:

Yes

Explanation:

The variables change in and experiment.

Alekssandra [29.7K]4 years ago
8 0

Answer:

If you are carefully enough to control everything, then everything that could change the result of your experiment won't happen.

Explanation:

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TRUE!!!!!!

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Which is a major area of study in psychology?mentalism totalculturalism behaviorism actionism
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Before the first behaviorist (Watson), Psychology was a part of philosophy.

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Look at the diagram below. From the frame of reference of the person riding
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3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
3 years ago
At a dock, a crane lifts a 2009 kg container 20 m, swings it out over the deck of a freighter, and lowers the container into the
deff fn [24]

Answer:

W = 157.5kJ

Explanation:

Assuming it moves the container at constant speed, the work done by the crane will be equal to the variation of the potential gratitational energy on the container:

Wc = \Delta E = m*g*(h2 - h1)  where h2= -8m  and  h1=0m

Wc = 157.5kJ

6 0
4 years ago
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