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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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A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6
steposvetlana [31]

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle.  There are two forces on the object: tension force T pulling up and weight force mg pulling down.

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∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle.  There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

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3 years ago
After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position
KiRa [710]

As we know by work energy theorem

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W_p = \frac{1}{2}mv_1^2 - 0

so we can say

W_p = \frac{1}{2}mv_1^2

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