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3 years ago

What were Katherine Johnson publications?

1 answer:

7 0

Answer: 1 Reaching for the Moon : the autobiography of NASA mathematician Katherine 2 Johnson by Katherine G Johnson( Book )

3 Katherine Johnson by Thea Feldman( Book )

4 Katherine Johnson by Ebony Wilkins( Book )

5Counting the stars by Lesa Cline-Ransome( Book )

6An Act to Award Congressional Gold Medals to Katherine Johnson and Dr.

Explanation:

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A person jogs 10 meters in 5 seconds calculate her avarage speed

Bas_tet [7]

Answer:

So first determine how many times does 5 go into 100.

100/5=20

Thus in 100 seconds the boy would have traveled a distance of 10 meters 20 times.

10*20=200

If the boy travels 10meters in 5 seconds, he travels 200m in 100seconds.

8 0

4 years ago

Read 2 more answers

A snail is at the bottom of a well, 115 feet deep. On day 1, it starts climbing up the side. That night it rests and slips down

julia-pushkina [17]

Answer:

Number of days needed = 57

Explanation:

Depth of well = 115 feet

Distance traveled in day = 3 feet up

Distance traveled in night = 1 feet down.

Total displacement in 1 day = 3 - 1 = 2 feet up

When the snail reaches 112 feet up, the next 3 feet up motion on day will make it reach the top.

$\texttt{Days to travel 112 feet up = }\frac{112}{2}=56$

After 56 days the displacement of snail is 112 feet.

On the next day it moves up by 3 feet and reaches out of well.

Number of days needed = 57

4 0

3 years ago

Charlotte is driving at 66.5 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 3

Alborosie

Answer:

the distance traveled by Charlotte in feet is 338.44 ft

Explanation:

Given;

speed of Charlotte, u = 66.5 mi/h

time of motion, t = 3.47 s

The distance traveled by Charlotte in feet is calculated as;

$Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft$

Therefore, the distance traveled by Charlotte in feet is 338.44 ft

7 0

3 years ago

Find the following answer based on the image.

saul85 [17]

We are given:

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

Finding g:

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² (approx)

5 0

3 years ago

So, what is the answer of this assignment?

Ket [755]

What assignment, you have to work with me here i don't know what were working on

6 0

3 years ago