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evablogger [386]
2 years ago
6

What is 6.9 × 107 written in standard form?

Physics
2 answers:
aalyn [17]2 years ago
8 0

Answer:

DDDDDD

Explanation:

n200080 [17]2 years ago
7 0
Answer: 6.9x 107 in standard form is 69,000,000

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True hope this helped
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IN WHAT WAY HAS TECHNOLOGY USED ABOARD THE INTERNATIONAL SPACE STATION BENEFITTED HUMANS BACK ON EARTH?
drek231 [11]

Answer

The ways are;

• By commercializing low earth orbit

• By supporting water purification efforts worldwide

• By bringing space station ultrasound equipment on earth

• By improving eye surgery with space hardware

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There is a revolutionary commercial pathways that is currently opening access to space.For example, NASA is purchasing commercial cargo resupply to space station which enable U.S business to develop a competition in selling services to others while freeing NASA resources for deep space exploration. Using  technology developed in space, the Water Security Corporation has deployed systems using NASA water-processing technology to ensure drinkable water availability for human survival. Medical care in remote regions is now  facilitated by use of small ultrasound units, tele-medicine  and remote guided methods similar to those used in space. The Eye Tracking Device experiment has enable researchers to have an insight into corrective laser surgeries.


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Leona [35]

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These tectonic plates rest upon the convecting mantle, which causes them to move. The movements of these plates can account for noticeable geologic events such as earthquakes, volcanic eruptions, and more subtle yet sublime events, like the building of mountains.

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3 years ago
An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res
murzikaleks [220]

Answer:

(a) V=11.86\ V

(b) V=9.76\ V

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

V=R.I

(a) The electromagnetic force of the battery is \varepsilon =12.6\ V and its internal resistance is R_i=0.06\ \Omega. Knowing the equivalent resistance of the headlights is R_e=5.2\ \Omega, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)

Solving for i

\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A

i=2.28\ A

The potential difference across the headlight  bulbs is

V=\varepsilon  -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V

V=11.86\ V

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

V=\varepsilon  -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V

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A parachute falling to the ground.<br><br>​
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a parachute falling to the ground is uniform

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