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Anastasy [175]
3 years ago
8

Why do stars start fusing helium into carbon and oxygen

Physics
1 answer:
nlexa [21]3 years ago
6 0

Answer:

Because they need more energy.

Explanation:

Stars fuse elements in the first place to get energy. In turn, this makes the elements heavier.

Hope this helped.

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A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 235 Hz. A person on the pl
zysi [14]

Answer: 2.9 m/s

Explanation:

The frequency of the beat is 4 Hz

The relative Doppler frequency is 235 + 4 = 239 Hz

We would be solving this question, using the formula for Doppler's effect

f(d) = f(v+vr)/(v-vs), where

F = 235 Hz

F(d) = 239 Hz

v = 344 m/s and vr = vs

239 = 235 (344 + vr) / (344 - vr)

239 ( 344 - vr) = 235 (344 + vr)

82216 - 239 vr = 80849 + 235 vr

82216 - 80849 = 235 vr + 239 vr

1376 = 474 vr

vr = 1376/474

vr = 2.9 m/s

Thus the speed the platform should move is 2.9 m/s

5 0
3 years ago
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marysya [2.9K]

Answer:1. INCREASE 2. DECREASE 3. DECREASE

Explanation:

3 0
3 years ago
The barrier to C-C bond rotation in iodoethane is 13 kJ/mol. What energy can you assign to an H-I eclipsing interaction?
n200080 [17]

Answer:

5J/mol

Explanation:

H-H eclipsed = Torsional strain = 4.0 kJ/mol

H-CH3 eclipsed = Mostly torsional strain = 6.0 kJ/mol

CH3-CH3 eclipsed =Torsional and steric strain = 11.0 kJ/mol

CH3-CH3 gauche = Steric strain = 3.8 kJ/mol  

You know each H-H eclipsed is 4 kJ/mol (8 kJ/mol total) and the reported barrier is 13 kJ/mol. Therefore the H-I eclipsed interaction must be 5 kJ/mol, (13-8 = 5).

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3 years ago
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"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This
LUCKY_DIMON [66]

Answer:

Imp = 5626.488\,\frac{kg\cdot m}{s}

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

a =\frac{781..25\,N}{(1.27\,s)^{2}}

a = 484 \frac{N}{s^{2}}

The equation is:

F_{x} = 484\,\frac{N}{s^{2}}\cdot t^{2}

The impulse done by the engine is given by the following integral:

Imp=484\,\frac{N}{s^{2}} \int\limits^{3.50\,s}_{2\,s} {t^{2}} \, dt

Imp = 161.333\,\frac{N}{s^{2}}\cdot [(3.50\,s)^{3}-(2\,s)^{3}]

Imp = 5626.488\,\frac{kg\cdot m}{s}

7 0
4 years ago
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