1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bogdanovich [222]
3 years ago
13

Aristarchus measured the angle between the Sun and the Moon when exactly half of the Moon was illuminated. He found this angle t

o be A greater than 90 degrees. B exactly 90 degrees. C less than 90 degrees by an amount too small for him to measure. D less than 90 degrees by an amount that was easy for him to measure.
Physics
1 answer:
frutty [35]3 years ago
8 0

Answer:

when the Sun illuminates half of the Moon it must be at 90°

Explanation:

The Moon has a circular motion around the Earth and the relative position of the sun, the earth and the moon create the lunar phases.

For this case when the Sun illuminates half of the Moon it must be at 90°, this angle changes with the movement of the moon, it is zero degree for the new moon and 180° for the full moon

You might be interested in
Lactic acid is produced during low-moderate exercise <br> True <br> False
Marizza181 [45]
The answer to your question is true.
3 0
3 years ago
Read 2 more answers
he electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
Paraphin [41]

Answer:

R₁ = 50.77 Ω

Explanation:

Since, we know that:

Electric Power = P = VI

but from Ohm's Law:

V = IR

(or) I = V/R

Therefore,

P = V²/R

(OR) R = V²/P

where,

V = Battery Voltage

R = Resistance of combination

FOR SERIES COMBINATION:

R = Rs = (57 V)²/48 W

Rs = 67.69 Ω

but, we know that:

Rs = R₁ + R₂

R₁ + R₂ = 67.69 Ω

R₁ = 67.69 Ω - R₂  __________ eqn (1)

FOR PARALLEL COMBINATION:

R = Rp = (57 V)²/256 W

Rp = 12.69 Ω

but, we know that:

Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω

using eqn (1) and value of R₁ + R₂, we get

Rp = 12.69  = R₂(67.69 - R₂)/67.69

859.08 = 67.69 R₂ - R₂²

R₂² - 67.69 R₂ + 859.08 = 0

Solving this quadratic equation we get the answers:

Either, R₂ = 50.76 Ω

Either, R₂ = 16.92 Ω

Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,

<u>R₂ = 16.92 Ω</u>

using this value in eqn (1), we get:

R₁ = 67.69 Ω - 16.92 Ω

<u>R₁ = 50.77 Ω</u>

4 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
3. Describe the flow of one molecule of water through the water cycle, beginning in the ocean.
scZoUnD [109]

molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.

6 0
3 years ago
Read 2 more answers
The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of plane
valkas [14]

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2

6 0
3 years ago
Other questions:
  • Where do seasons alternate between a very dry season and a monsoon season, with a stable warm climate?
    6·1 answer
  • What is true about all uranium atoms?they each have the same number of nuclear particles.they each have the same number of neutr
    15·2 answers
  • The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re
    13·1 answer
  • Is the
    15·1 answer
  • A grapefruit has a weight on earth of 4.9 newtons. what is the grapefruit's mass?
    8·1 answer
  • Which of the following correctly shows the chain of energy transfers that create surface currents on the ocean?
    11·1 answer
  • A 100kg car travels at 2m/s. what is the momentum in kg
    7·1 answer
  • PLEASE HELP ME!!! IREALLY NEED IT AND I WILL MARK BRAINLIST IF YOU ANSWER CORRECTLEY!!! mass of 0.22kg is oscillating on the pla
    5·2 answers
  • 200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as i
    13·1 answer
  • Define a dipole. Hence, write the expression for calculating the electric moment of a dipole​
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!