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2 years ago

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Which item is an example of a primary source? A. An academic paper analyzing John F. Kennedy's presidency B. A transcript of a p

residential speech C. A textbook about American foreign policy D. A newspaper article reflecting on the anniversary of the end of the Cold War

1 answer:

Viktor [21]2 years ago

7 0

A transcript of a presidential speech (APEX Class ;)

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Scheduling comes before sequencing. In scheduling, we develop plans for our work centers where production activities occur. We l

ser-zykov [4K]

Answer: 89

Explanation:

4 0

2 years ago

The aluminum atom______electrons to form an ion.<br> The ion that is formed is______.

Alchen [17]

The aluminum atom_loses_____electrons to form an ion.

The ion that is formed is_Al³⁺_____.

aluminium has the electronic configuration as 1s² 2s² 2p⁶ 3s² 3p¹

from the electronic configuration , we see that aluminium can attain stability by losing 3 electrons from outer shell.

after losing 3 electrons , the ion formed is given as Al³⁺

hence the correct options to fill in the blanks are lose and Al³⁺

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2 years ago

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As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti

mario62 [17]

Temperature stays the same

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2 years ago

Ivanshal [37]

Answer:

The velocity is 60 km/hr.

Explanation:

<h3><u>Given:</u></h3>

Displacement (d) = 480 km = 48000 m

Time (t) = 8 Hours = 480 minute

Velocity (v) = ?

Now,

Velocity = Displacement ÷ Time

v = d/t

v = 480/8

v = 60 km/hr

Thus, The velocity is 60 km/hr.

<u>-TheUnknownScientist 72</u>

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2 years ago

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average intensity of approximately

Dmitrij [34]

Intensity of sunlight at given position is defined as power received per unit area

so here we can say

$I = 2 kJ/s*m^2$

area on which photons are received is given as

$A = 4.80 cm^2 = 4.80 * 10^-4 m^2$

now we can find the power received due to sunlight

$P = I*A$

$P = 2* 10^3 * 4.80 * 10^-4$

$P = 0.96 Watt$

now we can say this power is due to photons that strikes on surface of earth

so here we can say

$P = N\frac{hc}{\lambda}$

given here that

$\lambda = 510 nm$

$0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}$

$0.96 = N * 3.88 * 10^{-19}$

$N = \frac{0.96}{3.88*10^{-19}}$

$N = 2.47 * 10^{18}$

so it will strike 2.47 * 10^18 photons on given area per second

3 0

3 years ago