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3 years ago

What discovery by galileo supported the helicentric (sun centered) model of our solar system?

1 answer:

4 0

Firstly the telescope helped with this discovery, in that he saw rings of Saturn and of Jupiter, which has moons orbiting around it as well as shadows of the moon and Mars and took very detailed notes on them seeing that shadows moved across the moon as seasons changing.

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Consider a coccus and a rod of equal volume. Which is more likely to survive in a dry environment? Explain your answer Which wou

Oxana [17]

Answer:

Coccus is suitable for dry environment and a rod is better adapted for moist environment.

Explanation:

The probability of Coccus would be better suitable for a environment which is dry. A sphere posses less surface area for the volume, as such, moisture which is lost through osmosis is less in dry environment conditions.

During a moist environment, the ratio of the surface area to volume of a rod shaped bacterium is high, which allow higher efficiency to transfer water and solutes into and outside of the cell, and to make possible for a cell which is in rod shaped, have a metabolic rate higher.

4 0

4 years ago

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at rrr = 4.0 mm is πr

max2010maxim [7]

Complete Question

A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.

Answer:

The phase at the second point is $\phi _2 = 1.57 \ rad$

Explanation:

From the question we are told that

The wavelength of the spherical wave is $\lambda = 2.0 \ mm = \frac{2}{1000} = 0.002 \ m$

The first radius is $r_1 = 4.0 \ mm = \frac{4}{1000} = 0.004 \ m$

The phase at that instant is $\phi _1 = \pi \ rad$

The second radius is $r_2 = 3.5 \ mm = \frac{3.5}{1000} = 0.0035 \ m$

Generally the phase difference is mathematically represented as

$\Delta \phi = \phi _2 - \phi _1$

this can also be expressed as

$\Delta \phi = \frac{2 \pi }{\lambda } (r_2 - r_1 )$

So we have that

$\phi _2 - \phi _1 = \frac{2 \pi }{\lambda } (r_2 - r_1 )$

substituting values

$\phi _2 - \pi = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )$

$\phi _2 = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) + 3.142$

$\phi _2 = 1.57 \ rad$

7 0

3 years ago

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

nignag [31]

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that

[tex]f=mgsin \theta [tex]

Therefore, option B is the correct option.

8 0

3 years ago

How do hurricanes differ from tornadoes?

denis-greek [22]

Answer:

tornadoes bxjndndnbdbbdbbzbbsbZbbhhhw

5 0

3 years ago

Based on the graph of kinetic energy given (gray curve in the graphing window), sketch a graph of the baseball's gravitational p

chubhunter [2.5K]

1) Physical principles:

a) Total mechanical energy = kinetic energy + potential energy.

b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)

2) Notation:

a) Total mechanical energy: ME.

b) Kinetic energy: KE

c) Gravitational potential energy: PE

∴ ME = KE + PE = constant

3) Solution:

a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.

b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0

c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).

d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.

At that point, KE = 0, and PE = ME.

That is the point of highest altitude and where the speed is zero.

d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.

Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.

The PE may be sketched on the same graph along with the KE and the ME.

The graph is attached. The red line is the ME and the blue line is the PE.

Note that at any point in the graph PE + KE = ME.

5 0

4 years ago

Read 2 more answers