All categories

3 years ago

Edwin Hubble calculated the expansion rate of the universe. On what evidence did he base his calculation?

2 answers:

7 0

Edwin Hubble calculated the expansion rate of the universe. The evidence that he base his calculation is the differences in redshift for galaxies. The answer is letter B. the red shift of galaxies was directly proportional to the distance of the galaxy from earth. It means that bodies farther away from Earth were moving away faster. The Hubble’s constant is the ratio of distance to redshift equal to 170 kilometers per second per light year of distance.

Anni [7]3 years ago

7 0

Answer: B

Explanation:

You might be interested in

How can you show positive body languages with your mouth

sergey [27]

By just smiling. It normally makes you look more happy.And if your happy that's positive body languge

7 0

3 years ago

Choose whether each of the following statements is true or false. In order to move a massive crate sitting on the floor, the for

velikii [3]

Answer:

Explanation:

1. False

The force you apply on crate is equal and opposite to the force that crate applies on you by Newton's third law of motion.

The force must over come the static frictional force between the crate and the floor.

2. True

The object can move along another direction than the direction of net force. For example, when a car slows down, the net force is opposite to the direction of motion.

3. True

An object moving at constant velocity has zero net force acting on it.

4. False

An object at rest has forces acting on it but the summation of all the forces is zero i.e. the net force is zero.

6 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

If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

Gre4nikov [31]

Answer: 500 s

Explanation:

Speed $v$ is defined as a relation between the distance $d$ and time $t$ :

$v=\frac{d}{t}$

Where:

$v=3(10)^{8}m/s$ is the speed of light in vacuum

$d=1.5(10)^{11}m$ is the distance between the Earth and Sun

$t$ is the time it takes to the light to travel the distance $d$

Isolating $t$ :

$t=\frac{d}{v}$

$t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}$

Finally:

$t=500 s$

5 0

3 years ago

An electron in a mercury atom drops

aksik [14]

Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.

<h3>How to calculate the photon energy?</h3>

In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.

Mathematically, the Planck-Einstein equation can be calculated by using this formula:

E = hf

<u>Where:</u>

h is Planck constant.

f is photon frequency.

In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.

Read more on photons here: brainly.com/question/9655595

#SPJ1

4 0

2 years ago