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natita [175]
4 years ago
15

Was Copernicus heliocentric or geocentric

Physics
2 answers:
Anuta_ua [19.1K]4 years ago
8 0
Copernicus preferred the heliocentric model.
Kitty [74]4 years ago
5 0
Copernicus was eccentric. He didn't accept the prevailing model of the solar system.
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if the displacement of a body is proportional to the square of time, state the nature of motion of the body ​
Liula [17]

Explanation:

If the displacement of an object is proportional to the square of the time taken then the body is moving with uniformly accelerated motion as it will follow Newton's second equation of motion for a particular initial velocity, which can be given by, s=ut+21at2.

hope this is helpful to you

3 0
3 years ago
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take t
marin [14]

Answer:

T=66262.4s

Explanation:

From the question we are told that:

Altitude A=2.90 *10^7

Mass m=5.97 * 10^{24} kg

Radius r=6.38 *10^6 m.

Generally the equation for Satellite Speed is mathematically given by

V=(\frac{GM}{d} )^{0.5}

V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}

V=3354.83m/s

Therefore

Period T is Given as

T=\frac{2 \pi *a}{V}

T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}

T=66262.4s

4 0
3 years ago
Solids that have flat sides and a repeating pattern of atoms are called ______________.
DiKsa [7]
 they are called crystals

7 0
4 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
HELP PLEASE!!!!!!!
Roman55 [17]

The silver substance is most likely a metal. Your answer is A.

8 0
3 years ago
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