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Fittoniya [83]
3 years ago
9

You have decided to study the effect of loud noise on plant growth. You put one plant in a quiet room and the other plant in you

r school's band hall. Providing each plant with the same amount of water is an example of a
a. data.
b. control.
c. variable.
d. observation.
Physics
2 answers:
-Dominant- [34]3 years ago
5 0

Answer:

The correct answer will be option-B.

Explanation:

The scientific experiments are performed to study the hypothesis and prove it.  The scientific factor to be studied in the experiment are known as the "variables". The variables can be "independent variables" which can control and changed, "dependent variables" which remains constant and to be studied and control variables which are kept constant in the experiment.

In the given question, the experiment is performed to study the effect of "loud noise" on the growth of plants so the dependent variable is plant growth dependent on loud noise. Since the same amount of water is provided to each plant will be a controlled variable.

Thus, Option-B is the correct answer.

wel3 years ago
4 0
The answer is B. Controlled variable.<span />
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Answer:

V = 25.3 , θf = 36.7° below the horizontal

Explanation:

Given Vo = 22m/s , θ = 23°

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Vy = VoSinθ – gt

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4 years ago
* A 5 kg ball is dropped from a height of 20 m. How much kinetic energy will it have 10 m above the ground? a. 490 J b. 392 J C.
Oksi-84 [34.3K]

Answer:

A. 490

Explanation:

soln

mass = m = 5kg

Height = h = 10m

Acceleration due to gravity = g = 9.8ms²

K.E = 1/2 × mass × (velocity)²

Recall from equations of motion

v² = u² + 2gh

Therefore,

K.E = 1/2 × mass × ( u² + 2gh)

K.E = 1/2 × 5 × ( 0² + 2×10×9.8)

K.E = 1/2 × 5 × 196

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6 0
2 years ago
In ce stare mecanica este un televizor fata de masina
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3 years ago
A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r
kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

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K.E = K.E=\frac{1}{2}I\omega^2

where,

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on substituting the values, we get

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3 years ago
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