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Vinvika [58]
3 years ago
11

it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m

oon. how fast did they travel?
Physics
2 answers:
Jet001 [13]3 years ago
7 0
We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values, 
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!
grin007 [14]3 years ago
6 0
Their speed was 5700 km/hour
you will change days to hours by multiplying by 24 and then add it to the 19.5 hours...you will divide the distance by the result to get the speed 
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With respect to the measure to be used in the competition, the appropriate SI unit is meter. This is the measure of length or distance covered.

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Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
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Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
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Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

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where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

6 0
3 years ago
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