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Gelneren [198K]
3 years ago
7

Emily wanted to see what battery lasted the longest. She put each battery into a clock and recorded the time that the clock stop

ped. She used Duracell, Energizer, Kroger brand, and EverReady. Iv,Dv, Control, consistent
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

This question is asking to identify the following variables:

Independent variable (IV): Battery

Dependent variable (DV): Time the clock stopped

Constant: Same clock

Control: No stated control

Explanation:

The independent variable in an experiment is the variable that is subject to manipulation or change by the experimenter. In this experiment, the independent variable is the BATTERIES (Duracell, Energizer, Kroger brand, EverReady).

The dependent variable is the variable that responds to the changes made to the independent variable. It is the variable that the experimenter measures. In this case, the dependent variable is the TIME IT TAKES FOR THE CLOCK TO STOP.

Constants or control variable is the variable that the experimenter keeps constant or unchanged for all groups throughout the experiment in order not to influence the outcome of the experiment. The constant in this case is the SAME CLOCK USED.

Control group is the group that does not receive the experimental treatment or independent variable in an experiment. In this case, all groups received a different kind of battery.

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What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?
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Read 2 more answers
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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