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Vedmedyk [2.9K]
2 years ago
11

How long does it take to reach the moon?

Physics
2 answers:
alexdok [17]2 years ago
5 0

Over the past few decades man has always embarked on different space exploration by traveling to the moon. A typical Journey to the moon from Earth will take about 3 days in recent times.

  • It takes about 3 days for a spacecraft to reach the Moon. During that time a spacecraft travels at least 240,000 miles

  • In the year 1969 the first man (Neil Armstrong ) landed on the moon, the journey took him 8 days, 3 hours, 18 minutes, 35 seconds, on 16 July.

Research on space exploration is still growing and expanding, with large amount of funding and support globally.

Learn more about Space and moon exploration:

brainly.com/question/3983943

aev [14]2 years ago
5 0

Answer:

about 3 days

thanks

have a great day

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3 years ago
skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular
laiz [17]

Answer:

L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

L=I\omega

Substitute all the values,

L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

So, its angular momentum is equal to 11.3\ kg-m^2/s.

8 0
3 years ago
Describe an object's velocity when an acceleration-time graph is zero?
svp [43]
Anything times zero is zero
7 0
3 years ago
Read 2 more answers
NEWTONS SECOND LAW LAB REPORT
hram777 [196]

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here this may help.

8 0
2 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
2 years ago
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