All categories

3 years ago

What causes the lunar phases we see on Earth? Give a few examples of different phases of the moon and what each looks like.

1 answer:

5 0

What causes us to see the moon is the Sun. The moon is usually above or below the Earth. So light from the Sun hits the moon and the moon reflects the light back. Some of it to Earth. For example, when a new moon happens the moon is in between the Sun and the Earth

SUN --------LIGHT-----------> MOON EARTH

What happens there is that none of the light gets reflected off the part of the moon that we can see from the ground, giving the illusion that it is not there.

During a full moon the Earth is in between the Sun and moon

SUN -------LIGHT---------> EARTH MOON

Now, nearly all the light from the Sun gets reflected off the surface of the moon and to us on Earth

You might be interested in

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

$Circular \; speed (V) = \frac {2 \pi r}{t}$

Where;

r represents the radius and t is the time.

Substituting into the formula, we have;

$Circular \; speed (V) = \frac {2*3.142*2600}{360}$

$Circular \; speed (V) = \frac {16338.4}{360}$

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

$Centripetal \; acceleration = \frac {V^{2}}{r}$

Where;

V is the circular speed (velocity) of an object.
r is the radius of circular path.

Substituting into the formula, we have;

$Centripetal \; acceleration = \frac {45.38^{2}}{2.6}$

$Centripetal \; acceleration = \frac {2059.34}{2600}$

Centripetal acceleration = 0.79 m/s²

3 0

3 years ago

Hi please please help me you’ll get points and I’ll give you brainliest

leonid [27]

Answer:

see below

Explanation:

Question 1)

57kg to ug

$126\:Kg(\frac{1*10^{9}ug }{1kg} )\\\\=126(10^{9} )\\\\=126,000,000,000 ug$ (there are 9 zeros)

Question 2)

126lbs to N

$126lbs(\frac{4.45N}{1lb} )\\\\=126(4.45N)\\\\=506.7N$

7 0

3 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

3 years ago

What time is the eclipse happening tonight?.

andrew11 [14]

The first location to see the partial solar eclipse begin is at 3.58 a.m. EST (08:58 UTC), the greatest point of total solar eclipse occurs at 6 a.m. EST (11:00 UTC) and the last location to see the partial eclipse end is at 8:02 a.m. EST (13:02 UTC) according to Time and Date.

3 0

2 years ago

A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o

Ierofanga [76]

On sources it says it would just be the super giant star

3 0

3 years ago