PHYSICS COLLEGE

Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....? Listed following are locations and times atwhich different phases of the Moon are visible from Earth’sNorthern Hemisphere. Match these to the appropriate moon phase.

The three given moon phases are:

Waxing Crescent Moon

Waning Crescent Moon

Full Moon

The things we need to match to the above three topicsare:

visible near western horizon an hour after sunset
rises about the same time the sun sets
visible near eastern horizon just before sun rises
occurs about 3 days before new moon (i know this is waningcrescent)
visible due south at midnite
occurs 14 days after new moon (i know this is full moon)

Answers

Answer 1

Answer:

Answer:

Explanation:

*visible near western horizon an hour aftersunset WAXING CRESCENT

*rises about the same time the sun sets FULLMOON

*visible near eastern horizon just before sun rises WANING CRESCENT

*occurs about 3 days before new moon WANING CRESCENT

*visible due south at midnite FULL MOON

*occurs 14 days after new moon FULL MOON

Related Questions

Which of the following is a TRUE statement? a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible. b. It is not possible to convert work entirely into heat. c. The second law of thermodynamics is a consequence of the first law of thermodynamics. d. It is impossible to transfer heat from a cooler to a hotter body. e. All of these statements are false.
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At a specific temperature the pressure is 102.1 kPa at sea level and 87.8 kPa at h = 1,000 m. (Round your answers to one decimal place.) (a) What is the pressure at an altitude of 4500 m? kPa (b) What is the pressure at the top of a mountain that is 6165 m high?
The most interpersonal constructive passion response to relational conflict is..
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?
Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

A 60kg woman on skates throws a 3.9kg ball with a velocity of37m.s west. What is the velocity of the woman?

Answers

Answer:

2.405 m/s

Explanation:

Given that,

Mass of a women, m₁ = 60 kg

Mass of a ball, m₂ = 3.9 kg

Velocity of the ball, v₂ = 37 m/s

We need to find the velocity of the woman. It is a concept based on the conservation of linear momentum. Let v₁ is the velocity of the woman. So,

$m_1v_1=m_2v_2\n\nv_1=(m_2v_2)/(m_1)\n\nv_1=(3.9* 37)/(60)\n\nv_1=2.405\ m/s$

So, the velocity of the woman is 2.405 m/s.

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

Chuck wants to investigate how gas moleculesmove in a container. Which model would be most
helpful to represent this motion?
A. stacking blocks to build a tower
B. freezing water in an ice cube tray
C. bouncing elastic balls off of each other and
the walls of a room
D. placing a closed, water-filled plastic bag in
the sun and watching condensation form

Answers

The answers C the molecules in gas move rapidly and all around they are spread out and bounce off each-other

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

A car (m = 2000 kg) is going around an unbanked curve at the recommended speed of 11m/s (24.6 MPH). (a) If the radius of the curvature of the path is 25m and the coefficient of static friction between the rubber tires and the road is µs = 0.70, does the car skid as it goes around the curve? (b) What will happen if the driver ignores the highway speed limit sign and travels at 18 m/s (40.3 MPH)? (c) What speed is safe for traveling around the curve if the road surface is wet from a recent rainstorm and the coefficient of static friction between the wet roud and the rubber tires is µs = 0.50?

Answers

Answer:

a) car does not skid, b) car skids, c) v = 11.07 m / s

Explanation:

a) When the car around in a curve all force must be exerted by friction, write Newton's second Law

Y axis (vertical)

N - W = 0

N = W = mg

X axis (radial

F = m a

The acceleration is centripetal

a = v² / r

fr = μ N

Let's calculate the maximum friction force

fr = μ m g

fr = 0.70 2000 9.8

fr = 13720 N

Let's calculate the force necessary to take the curve

F = m v² / r

F = 2000 11²/25

F = 9680 N

When examining these two values we see that the maximum value of the friction force is greater than the force to stay in the curve, for which the car does not skid

b) The speed of the driver is v = 18m / s, let's calculate the force to stay in the curve

F = 2000 18²/25

F = 25920 N

This force is greater than the maximum friction force, so it is a skating car

c) The friction coefficient decreases to μ = 0.5

fr = m a

μ mg = m v² / r

v = √μ g r

v = √(0.50 9.8 25)

v = 11.07 m / s

This is the maximum speed

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