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Alex_Xolod [135]

3 years ago

13

When using science to investigate physical phenomena, which characteristic of the event must exist?

1 answer:

Vesna [10]3 years ago

6 0

The correct answer is B. Measurable

Explanation:

The use of science to investigate a phenomenon implies using measurements and observations to better understand a phenomenon or test a hypothesis. Moreover, science focuses on natural phenomena that can be objectively studied through measurement instruments such as a thermometer, balance, hydrometer, etc.

In this context, for a phenomenon to be studied by science this needs to be measurable because the use of precise instruments as wells as numbers allow scientist to analyze and understand a phenomenon. Moreover, phenomena that depend on personal perspectives and cannot be measure is considered as non-scientific.

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Monica [59]

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The answer to your question is D An annual weather pattern

Explanation:

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3 years ago

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How long would it take light from the sun to reach mercury?

BARSIC [14]

About 499.0 seconds.....

7 0

4 years ago

the speed of a train is decreased in a uniform rate from 96 km/h to 48km/h through a distance of 800m. calculate the distance co

GREYUIT [131]

Answer:

1066.67 m

Explanation:

Given:

v₀ = 96 km/h = 26.67 m/s

v = 48 km/h = 13.33 m/s

Δx = 800 m

Find: a

v² = v₀² + 2aΔx

(13.33 m/s)² = (26.67 m/s)² + 2a (800 m)

a = -0.333 m/s²

Given:

v₀ = 26.67 m/s

v = 0 m/s

a = -0.333 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (26.67 m/s)² + 2 (-0.333 m/s²) Δx

Δx = 1066.67 m

Round as needed.

7 0

3 years ago

In an equation f = l^2-d^2/4l the intercept is<br>

DanielleElmas [232]

Answer:

the intercept is the orgin (0,0)

5 0

3 years ago

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di

Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block $\omega = 50 \ rad/s$

Disk radius (r) = 0.2 m

The block Initial velocity is:

$v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s$

Change in the block's angular speed is:

$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

However, on the disk, moment of inertIa is:

$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

The time t = 10s

∴

Frictional torques by the wall on the disk is:

$T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m$

Finally, the frictional force is calculated as:

$F = \dfrac{T}r{}$

$F= \dfrac{0.6}{0.2} \\ \\ F = 3N$

8 0

3 years ago