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Tatiana [17]
3 years ago
14

Describe the steps you use during an experiment

Physics
2 answers:
Nezavi [6.7K]3 years ago
5 0
Gather your materials
Do the experiment
Collect the data
DochEvi [55]3 years ago
4 0
<span>come up with a question you want answered, create a hypothesis, perform the experiment to test your hypothesis. obtain your data and analysis. write down your results and compare them to your hypothesis. then come up with a conclusion whether your hypothesis was right or wrong.</span>
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What does Einstein's famous equation for nuclear energy, E = mc^2, mean?
Nezavi [6.7K]
Yup the correct answer is A jus finished the quiz :)

hope i helped^_^

8 0
4 years ago
A wave with a period of 0.008 second has a frequency of?
tatyana61 [14]
The formula for frequency is f = 1/T where f is frequency and T is period in seconds. 
You have you period which is 0.008s and that is all you will need to solve or frequency in a wave:
f = 1/2
f = 1/0.008s
f = 125Hz
6 0
3 years ago
Read 2 more answers
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
Infant car seats are made to face the rear of the car. This is safer in a front end collision because of Newton's First law. New
kicyunya [14]
The answer is D hope it helps:)
5 0
3 years ago
Read 2 more answers
The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

M_e = mass of earth

Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

r_2 = 1 AU

T_2 = 1 year

M_s = mass of Sun

now we have

\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}

12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

4 0
3 years ago
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