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4 years ago

In which position are the earth, moon, and sun during a full moon?

Physics

1 answer:

Arte-miy333 [17]4 years ago

5 0

Answer:

The answer is B, although technically that is an eclipse.

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On a hot day, the deck of a small ship reaches a temperature of 48

AlekseyPX

The final temperature of the seawater-deck system is 990°C.

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

brainly.com/question/13860901

#SPJ1

6 0

2 years ago

If a melon has a a mass of 1 kg, how much does the melon weigh?

lesya [120]

Answer:

A. 10 N

Explanation:

weight of melon is 1×10=10N..........

6 0

3 years ago

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line

Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm

5 0

3 years ago

Am I the only one or do I hate scooping the cat box if you guys have cats.

LenaWriter [7]

THIS DUDE REALLY SHOWED US CAT TURDS...... IM DYING LM.AOOOOOOOO
BAHAHAHA

4 0

3 years ago

Read 2 more answers

Two projectiles are in flight at the same time. The acceleration of one relative to the other: A. is always 9.8 m/s2 B. can be a

zepelin [54]

Answer:

D. Is Zero

Explanation:

The acceleration on projectiles travelling through the air is only caused by gravity. This equal 9.81 m/s^2.

Other than gravity, there is no source of acceleration. Only air resistance causes projectiles of different surface areas to slow down at different rates.

As friction is not considered here (because no statement about surface areas of projectiles is given in the question), we do not need to consider it's effects and thus ONLY GRAVITY MATTERS.

Since gravity acting on both objects is equal, they have no acceleration relative to one another.

5 0

3 years ago