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rusak2 [61]
3 years ago
7

Hiaaa I need help pwease" concerning the planetary orbits why was Copernicus somewhat inaccurate?"

Physics
1 answer:
Yuki888 [10]3 years ago
4 0
Copernicus's heliocentric theory stated that the planets orbited in perfectly circular orbits. In reality, they are shaped like ellipses (ovals).
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If a stress of 1.05 × 107 n/m2 is applied to his ligament, how much will it stretch? the young\'s modulus for the acl is 1.10 ×
Ainat [17]

As we know that

Young's\: Modulus = \frac{stress}{strain}

here we know that

Young's Modulus = 1.10 \times 10^8

stress = 1.05 \times 10^7

now by the above formula we will have

1.10 \times 10^8 = \frac{1.05 \times 10^7}{strain}

strain = \frac{1.05 \times 10^7}{1.10 \times 10^8}

strain = 0.095


3 0
3 years ago
2. What is the percent composition of sulfur in H2SO4?
allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

6 0
3 years ago
Where does most of the oxygen at earth reside?
Lady_Fox [76]
Well most of the earth's oxygen resides in mineral oxides of the crust. a small fraction resides in the atmosphere and an even smaller fraction resides in the biosphere. still the biosphere is crucial to understanding the atmospheric oxygen budget as it controls short term exchanges between sediments and the atmosphere.
4 0
3 years ago
What is the current through an electric heater with a resistance of 38  when the potential difference is 240 V?
Zinaida [17]

\bf{ \underline{Given:- }}

\sf•  \: Resistance \:  (R)  \: = 38 \:  Ω

• \sf \:  Potential  \: difference \:  (V)   \: = 240  \: v

\bf{ \underline{To \:  Find:- }}

• \sf  \: The \:  current \:  through \:  an \:  electric  \: heater.

<h2>\bf{ \underline{ Solution :-}}</h2>

\bf  \red{•\: The  \: formula  \: of \:  Current  \: (I) =  \frac{V}{R}}

\sf \rightarrow I =  \frac{240}{38}

\sf \rightarrow I = 6.32

\sf \pink{Answer :-  \: The \:  current \:  through \:  an \:  electric  \: heater \: is \: 6.32 \: A.}

6 0
3 years ago
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

3 0
4 years ago
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