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3 years ago

What is the difference between the geocentric in the hello centric model of the solar system

Physics

1 answer:

xeze [42]3 years ago

6 0

Answer:

"Geocentric means the Earth is at the center. The prefix geo means of the Earth. ... The heliocentric model puts the sun at the center of the solar system."

Explanation:

Geocentric means earth is at the center and that is why Geocentric has Geo and Geo means earth And Heliocentric is puts the SUN in the center of the solar system

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A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$ ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$ ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m

7 0

4 years ago

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago

The area of a circular trampoline is 108.94 square feet. What is the radius of the trampoline? Round to the nearest hundredth.

djverab [1.8K]

The correct answer to this question is this one: B. 5.89 feet
<span>The area of a circular trampoline is 108.94 square feet. The radius of the trampoline is 5.89 feet.
</span>
For a circle,
A = (pi)r^2
108.94 ft^2 = (pi)r^2
r^2 = (108.94 ft^2)/(pi)
r = sqrt(108.94/pi) ft
r = 5. 89 ft

8 0

3 years ago

Read 2 more answers

IF you are in Space and push a bowling what happens to you and the bowling ball?

Anastaziya [24]

Answer:

The bowling ball did not change size or shape- the only thing that changed was the amount of gravity that pulls on it. But the mass of the bowling ball would never change. A bowling ball with a mass of 12 pounds on earth will have the mass of 12 pounds on the moon! Mass is the amount of atoms that a space fills.

Explanation:

I hope this helps! :D

4 0

3 years ago

Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume

REY [17]

Explanation:

The given data is as follows.

mass, m = 75 g

$T_{1} = 0^{o}C$

$T_{2} = 27^{o}C$

Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

= -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.

4 0

4 years ago