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Neko [114]
3 years ago
11

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

1.55 × 105 m. Suppose further that a second group of engineers programmed the orbiter to go to 1.55 × 105 ft. What was the difference in kilometers between the two altitudes? How low did the probe go?
Physics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

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lora16 [44]
<h3>Solution:</h3>

firstly solve parenthesis bracket ( )

-4×3 + 4×5x ≥ -6x + 9

-12 + 20x ≥ -6x + 9

Add 12 on both sides

-12 + 12 + 20x ≥ -6x + 9 + 12

20x ≥ -6x + 21

Add 6x on both sides

6x+ 20x ≥ -6x + 6x + 21

26x ≥  21

divide both side by 26

26 x/26 ≥21/26

x ≥21/26

3 0
3 years ago
An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0
Elina [12.6K]

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

3 0
2 years ago
Which of the following definitions describes a receptacle ground fault circuit interrupter (GFCI)?
jolli1 [7]

There are no appropriate definitions on the
list of choices that you've included.

And what does "the following" mean anyway ?
6 0
3 years ago
Ms. Mayo challenged her students to build a pendulum that would hit a block of wood and make it travel the farthest distance. Th
cestrela7 [59]

Answer:

<u>B) mass of pendulum bob</u>

4 0
2 years ago
The energy released from condensation in thunderstorms can be very large. Calculate the energy (in J) released into the atmosphe
makkiz [27]

Answer:

7.19 * 10^14J

Explanation:

Given that

Density of water Pwater= 1000kg/m3

R=2.1km = 2.1*10^3m

H= 2.3cm. = 2.3*10^-2m

Lv water= 2256 * 10^3J/kg

First, mass of water need to be calculated, using an imaginary cylinder

Density= Mass /Volume

Mass= Density* Volume

Volume of a cylinder= πR2h

Therefore mass= PπR2H

= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)

= 3.18 *10^8

Heat Released Qv = mLV

= 3.18*10^8 * 2236*10^3

= 7.19 * 10^14J

7 0
3 years ago
Read 2 more answers
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