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4 years ago

Make the following prefix conversions. 600,000 cm= ?? km i need help

Physics

1 answer:

Sati [7]4 years ago

4 0

Answer:

6 km

Explanation:

A kilometer is 100,000 centimeters, so 600,000 centimeters equals 6 kilometers.

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A current of 0.8 A flows through the motor for 3 minutes. Calculate the total charge in this time.

mixas84 [53]

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The given terms are :

Current = 0.8 Amperes

Time = 3 minutes = 3 × 60 = 180 seconds

we know that,

$current = \dfrac{charge}{time}$

now, let's solve for charge (q) :

$0.8 = \dfrac{q}{180}$

$q = 180 \times 0.8$

$q = 144 \:$

Total Charge (q) = 144 Coulombs

7 0

3 years ago

Define Work.<br>Thank You!

Anna35 [415]

Answer:

Activity in which one exerts strength or faculties to do or perform something. A job is an activity that a person engages in regularly to earn a livelihood.

<u>you are welcome!</u>

6 0

3 years ago

Read 2 more answers

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

4 years ago

You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over

Cerrena [4.2K]

Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

$\frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1}$

8 0

3 years ago

How long would it take to drive to alpha centauri (4.4 light-years away)?

oee [108]

A light year is a unit of length that is used to espress distances that are astronomical. It is equivalent to 5.9 trillion miles or 9.5 quadrillion meters. Light year is defined by the distance a beam of light would travel in one year. One light year is equivalent to the distance that light would travel in vacuum on one Julian year or 365 days as defined by the IAU or the International Astronomical Union. So, one light year is traveled for 1 year here on Earth. So, in order to drive to the alpha centauri, you would take about 4.4 years to do so.

6 0

3 years ago