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3 years ago

How did Galileo increase public support for Copernicus’s model?

Physics

2 answers:

miskamm [114]3 years ago

8 0

I think by using data collected by Tycho Brahe

just olya [345]3 years ago

3 0

Answer:

Explanation:

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Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a

Alex Ar [27]

A) Vector A

The x-component of a vector can be found by using the formula

$v_x = v cos \theta$

where

v is the magnitude of the vector

$\theta$ is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its x-component is

$A_x = A cos \theta_A = (50) cos 0^{\circ}=50$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ (negative since it is below the x-axis), so the x-component is

$B_x = B cos \theta_B = (120) cos (-70^{\circ})=41$

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

$v_y = v sin \theta$

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so $\theta_A = 0^{\circ}$ . So its y-component is

$A_y = A sin \theta_A = (50) sin 0^{\circ}=0$

- Vector B has a magnitude of 120 units and the direction is $\theta_B = -70^{\circ}$ , so the y-component is

$B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7$

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

8 0

3 years ago

A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s

boyakko [2]

Answer:

what diagram. I can't see any

5 0

3 years ago

A student of mass 40kg takes 10s to run up a flight of 50steps. If each step is 15cm high, calculate the Potential Energy of the

astra-53 [7]

Answer:

2943 J

Explanation:

Potential energy = mass x acceleration due to gravity x height in meters

(P.E. = mgh)

Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

PE = 40 x 9.81 x (0.15x50)

PE = 2943 J

3 0

4 years ago

Is that statement true or false?

Softa [21]

False, they don’t use the lures to produce light. Instead they use a strong flash of bioluminescence to scare off predators

3 0

3 years ago

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

Vinil7 [7]

Answer:

The distance is $r = 55430496 \ m$

Explanation:

From the question we are told that

The period of the moon $T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s$

The mass of the planet is $m_p = 9.38*10^{24} kg$

Generally the period of the moon is mathematically represented as

$T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

Here G is the gravitational constant with value

$G = 6.67 *10^{-11} \ N \cdot m^2/kg^2$

=> $T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

=> $103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }$

=> $272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}$

=> $r = \sqrt[3]{ 1.7031241*10^{23}}$ j

=> $r = 55430496 \ m$

8 0

3 years ago