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Dafna11 [192]
2 years ago
15

Which of the following occurs during the evolution of a star?

Physics
1 answer:
UNO [17]2 years ago
5 0
B
Hope this helps! ^-^
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Please help me anything will help
spin [16.1K]
Your answer is C) The speed of sound is higher in solids than in liquids. 
6 0
3 years ago
A new restaurant is interested in determining the best time-temperature combination for roasting a five-pound cut of lamb. The t
Leto [7]

Answer:

C

Explanation:

(c) The two cuts that are being roasted for each time-temperature combination are an example of replication.

In the question it is given that  From 10 identical cuts of lamb, 2 are randomly selected to roast using each of the time-temperature combinations in the same oven. Here it is an act of copying the exact sahpe size of the lamb in all cuts, which is nothing but replication. Moreover, this replication can help in proper comparision.

7 0
3 years ago
To ya is modeling the discovery of electromagnetic induction. Which procedures should she use
Vedmedyk [2.9K]

Answer:

moving a magnet into a coil of wire in a closed circuit.

Ed 2020

8 0
3 years ago
Calculate delta g for the reaction if the partial pressures of the initial mixture are pcl5 = .0029 atm, pcl3 = .27 atm, and pcl
AnnyKZ [126]

Answer: equation for the reaction is given below

PCL2+PCL3=PCL5

Where pcl2=0.40atm,pcl3=0.27atm

Pcl5=0.0029atm

Using ∆G=-RTin(PCL5/PCl2*PCL3)

Where R=8.314J/K/mol and T=298K

∆G=-8.314*298in(0.0029/0.40*.27)

∆G=8962.6J/mol

Explanation:

7 0
3 years ago
An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi
Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0
3 years ago
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