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iragen [17]
2 years ago
7

HELP PLEASE!!!!!!!!!!!!!!!

Physics
1 answer:
vovikov84 [41]2 years ago
7 0
The answer will be : B
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Why does data need to be reliable
Karolina [17]
So results can be shared and used by other scientists that want to use or replicate your experiment.
8 0
3 years ago
Read 2 more answers
Bob wanted to investigate the effects of plant fertilizer. He bought three identical plants and gave plant 1 fertilizer every mo
olga_2 [115]

Answer:

TIME he applied the fertilizer to each plant

Explanation:

Independent variable in an experiment is the variable that is subject to change or manipulation by the experimenter. In this experiment, Bob wanted to investigate the effects of plant fertilizer. Bob sets up the experiment by applying the fertilizer to each plant at DIFFERENT TIMES i.e. plant 1-every morning, plant 2-once a week, plant 3-never.

Based on this, it is obvious that the independent or manipulated variable is the TIME at which he applied the fertilizer. On the other hand, the dependent or measured variable is the height of the plants.

8 0
3 years ago
Which of the following regions of the electromagnetic spectrum has the highest energy
garri49 [273]

Answer:

B. Radiowaves

Explanation:

Radiowaves are on the higher end of the spectrum, then infrared, visible light, and UV light (which is the lowest energy)

6 0
2 years ago
A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.
zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius  r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is  ∝ v²

so we can say

\frac{a2}{a1} = \frac{v2}{v1}    .....................1

put here value

\frac{2}{1} = \frac{v2}{22.1}  

solve it

v2 = \sqrt{2 } × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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