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ziro4ka [17]
3 years ago
10

After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that

the question has already been asked and answered by several investigations. What should the scientist do?
The scientist should test the question anyway.
The scientist should broaden the scope of the question.
The scientist should come up with a new way to test the question.
The scientist should revise the question to address a gap in knowledge.
Physics
2 answers:
frozen [14]3 years ago
8 0

Answer: The scientist should revise the question to address a gap in knowledge.

Explanation:

The scientist should revise the question to address a gap in knowledge. is the correct option this is because of the fact that on revising the question as well as observing the findings of the other scientists the scientist may realize the fact that what other variable can be used for the purpose of the experiment and what other factor can be included in framing the question. This will help in evolution of the same findings by many researchers. This will add up more knowledge in the overall research procedure.

telo118 [61]3 years ago
6 0

Answer: Hello there!

If the question has been answered by several investigators, there is a high chance that there is already a good model that describes the moth.

But anyway, the scientist should revise the question to address a gap in knowledge, and see if he/her can do something to improve the knowledge in this particular subject.

If the scientist don't find anything that can be improved, there is no actual reason to lost time and resources in test something that has already been tested and is fully described (actually is useful to test something a lot of times, to see if different investigators obtain the same results, but in this case this question already has been answered by many investigators)

so the right answer is: "The scientist should revise the question to address a gap in knowledge"

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

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3 years ago
A driver brings a car to a full stop in 2.0 s. If the car was initially
vagabundo [1.1K]

Answer:

\boxed {\boxed {\sf D. \  -11 \ m/s^2}}

Explanation:

Acceleration can be found by dividing the change in velocity by the time.

a=\frac{v_f-v_i}{t}

where Vf is the final velocity, Vi is the initial velocity, and t is the time.

Since the car came to a complete stop, it's final velocity was 0 meters per second.

The initial velocity was 22 meters per second.

The time was 2.0 seconds.

v_f=0 \ m/s\\v_i=22 \ m/s\\t= 2.0 \ s

Substitute the values into the formula.

a=\frac{0 \ m/s-22 \ m/s}{ 2.0 \s }

Subtract in the numerator first.

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a=\frac{-22 \ m/s}{2.0 \ s}

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a= -11 \ m/s^2

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The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

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The velocity at this temperature is calculated as -

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Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

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