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Liono4ka [1.6K]
4 years ago
6

After watching a show about submarines, Jamil wants to learn more about the oceans. Which question could be answered through sci

entific investigation?
Which ocean has the best beaches?

Which type of whale is the most popular?

What ocean has the best scuba diving spots?

What substances dissolve in ocean water?
Physics
2 answers:
omeli [17]4 years ago
7 0

Answer: What substances dissolve in ocean water?

Explanation:

A scientific investigation can be define as the fact finding inquiry in which answers to the questions related to science can be determined. The scientific investigation can be conducted using scientific methodologies and experimental procedures.

What substances dissolve in ocean water? is the correct question which can be answered using experimental trials and observations.

 

lyudmila [28]4 years ago
6 0
He can ask what substances dissolve in ocean water. 
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4 years ago
Please help me I really need fast answer
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3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

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