Which part of the comet exists when it is away from the Sun? A. tail B. coma C. nucleus D. meteor E. core

cestrela7 [59]

Skip it but most probably the second one

3 0

4 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

3 years ago

As a recreational boat operator, what actions must you take when in a narrow channel?

alexandr402 [8]

One should never anchor in a narrow channel, until unless required very importantly. One should stay to the starboard side, and use a prolonged blast. The announcement must be done to alarm the nearby vessels, about your approach. The vessel should be kept at the outer limit of the starboard side.

6 0

3 years ago

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Help meeeee!!! ⇓⇓⇓⇓⇓

Dafna11 [192]

I believe you are correct, it is B: Diagnostic Services.

Diagnostic services are services like the staff at hospitals and the people who run machines that are related to medical needs.

If this is incorrect, please, don't refrain to tell me.

6 0

3 years ago

Read 2 more answers

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting: