The speed of light is 3 x 10 m/s.

To test the quality of water sources in her area, Shameka collects five samples from each water source and tests the quality of

xenn [34]

The missing part of the incomplete question is given below:

Which important step of scientific design is Shameka conducting?

repetition

replication

verification of results

using controlled variables

Answer:

Verification of results

Explanation:

The way toward gathering five examples of water from various sources is conveyed to confirm the outcome. By gathering water from five distinct areas of a similar source the analyst can genuinely find out the nature of the water in her region of remain.

On the off chance that after examples are tried it is found the water isn't sound, the outcomes would be acknowledged as it has been appropriately checked and a proper move would be made.

Thus, the correct answer is - verification of results

6 0

3 years ago

A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

$Xc = \frac{1}{2\pi fC}$

where,

$Xc\\$ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

$Xc$ =?

$Xc' = \frac{1}{2\pi f'C}$

$= \frac{1}{2\pi 2f C}$

$= \frac{1}{2} (\frac{1}{2\pi fC} )\\$

$Xc' = \frac{1}{2} Xc$

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

The capacitive reactance is doubled.
The capacitive reactance is traduced by a factor of 4.
The capacitive reactance remains constant.
The capacitive reactance is quadrupled.
The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

3 0

2 years ago

1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat

navik [9.2K]

Answer:

technically yes

Explanation:

with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...

4 0

3 years ago

Read 2 more answers

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

1.The distance moved by objects in a given interval of time can help us to decide which one is faster or slower. Do you agree wi

shutvik [7]

Answer:

The rate of change of distance is defined as speed.

Explanation:

The speed is defined as the rate of change of distance.

Speed = distance/ time

When we know the distance and the time, we get the value of speed. So, e know that who is moving fast or slow.

hen a graph is pltted beteen the distance and time, the slope of the graph gives the value of speed. So, by checking the slopes, hoseslope ismore, the speed is more and thusit is moving faster.

So, i agree with the statement.

6 0

3 years ago