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Helen [10]
2 years ago
7

Due today HELP HELP HELP

Physics
1 answer:
Anna [14]2 years ago
5 0

Vas happenin!

Independent variable : amount of water each day

Dependent variable: water on the windsill

Hypotheses: Ben wants to try by adding water each day to two different places. Will that work? Will that effect the water?


Hope this helps you out

*smiles*


-Zayn Malik
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What does the atomic number respresent?<br><br> (From my science homework)
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It represents where it is located in the periodic table and how many protons a element has. 
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At what height from the surface of the earth does the value of acceleration due to gravity be 2.45 m/s square where the radius o
Citrus2011 [14]

Answer:

Explanation: RADIUS OF EARTH = 6400X1000m =

ACC DUE GRAVITY ABOVE SURFACE OF EARTH = g' =2.45 m/s^2

ACC DUE GRAVITY ON SURFACE OF EARTH =g= 9.8 m/s^2

A/C TO FORmULA

g'/g=1-2h/Re

g'/g +2h/Re = 1

2h/Re =1- g'/g

2h= (1- g'/g)Re

2h=(1-2.45 /9.8)

6400X1000

2h = (0.75)6400X1000

2h = 4800000

h= 2400000

m

3 0
1 year ago
Please help with this
frosja888 [35]
912.

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3 years ago
Which scientist invented a model of the atom that most closely resembles the modern electron cloud model
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4 0
2 years ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
2 years ago
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