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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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Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air

Aleonysh [2.5K]

Answer:

a) I = 0 N s, b) v = -3.935 m / s, c) vf = 3.935 m / s, d) y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

I = ∫ (9200 t - 11500 t2) dt

I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

I = 9200 (0.8² -0) - 11500 (0.8³ -0)

I = 5888 -5888

I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

v² = v₀² - 2g y

v = √ (0 - 2 9.8 (-0.79))

v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

I = ΔP

I = m vf - m v₀

vf = (I + m v₀) / m

This is the impulse of women on the floor

vf = ( 0 + 68 (3.935)) / 68

vf = 3.935 m / s

d) let's use kinematics

v₂ = v₀² - 2gy

0 = v₀² - 2gy

y = v₀² / 2g

y = 3.935²/2 9.8

y = 0.790 m

8 0

3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

What are the differences between concave lens and concave mirrors?

Pavel [41]

Explanation:

A concave mirror can form real, inverted images of various sizes and virtual, erect and enlarged images whereas a concave lens forms only virtual, errect and diminished images.

8 0

3 years ago

What are the reactants of cellular respiration? Check all that apply. carbon dioxide energy glucose oxygen water

Rufina [12.5K]

Answer:

Glucose and Oxygen

Explanation:

Cellular respiration is the process whereby cells derives energy by the use of glucose and oxygen.

Organisms that use cellular respiration to produce their energy are known as heterotophs. They derive the glucose from food materials obtained from plant sources. They use the oxygen from the environment to liberate energy from the glucose obtained from feeding on plant materials.

Cellular respiration can be simply expressed as shown below:

GLUCOSE + OXYGEN → CO₂ + H₂O + ATP

The reactants are glucose and oxygen.

The products are CO₂, water and ATP

7 0

3 years ago

Read 2 more answers

How many grams of ice at 0°c will melt if 1.50 kj of heat are added? (the molar heat of fusion of water is 6.01 kj/mol.) 4.49 g

Nata [24]

The computation would be:moles = mass/ Molar Mass, but we are looking for the mass, so rearranging, will give us: mass = moles x MM
Q = moles x Hf
Q = (mass/MM) x Hf
mass = (Q x MM) / Hf
= (1.50-kJ x 18.0-g/mol) / 6.01-kJ/mol
=4.49 g H20 is the answer

7 0

3 years ago

Read 2 more answers