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3 years ago

What moon phase occurs 3-4 days after a waning gibbous?

Physics

1 answer:

rewona [7]3 years ago

8 0

The "waning gibbous" begins immediately after the Full Moon and lasts about a week.

At the end of the week, the moon is no longer gibbous, but it's still waning. At the instant it's exactly half-illuminated, it's called "Third Quarter", and then it continuous to wane for another week.

So 3 to 4 days after the END of the waning gibbous phase, it's a Waning Crescent. It still has another 3 to 4 days of waning to go, before it wanes away to nothing and we have the next New Moon.

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1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the

Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = $\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}$

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= $\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}$

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0

3 years ago

A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars

VLD [36.1K]

In order to compute the final velocity of the trains, we may apply the principle of conservation of momentum which is:
initial momentum = final momentum
m₁v₁ = m₂v₂

The final mass of the trains will be:
10,000 + 10,000 = 20,000 kg
Substituting the values into the equation:

10,000 * 3 = 20,000 * v
v = 1.5 m/s

The final velocity of the trains will be 1.5 m/s

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3 years ago

Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in

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Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

$t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C$

Therefore, the final temperature of the tea is 7.39⁰C.

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3 years ago

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Answer:

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3 years ago