Which is the greatest contribution of Gregor Mendel to society?

The orbital radius of the Earth (from Earth to Sun) is 1.496 x 10^11 m.

mrs_skeptik [129]

Explanation:

The orbital radius of the Earth is $r_1=1.496\times 10^{11}\ m$

The orbital radius of the Mercury is $r_2=5.79 \times 10^{10}\ m$

The orbital radius of the Pluto is $r_3=5.91 \times 10^{12}\ m$

We need to find the time required for light to travel from the Sun to each of the three planets.

(a) For Sun -Earth,

Kepler's third law :

$T_1^2=\dfrac{4\pi ^2}{GM}r_1^3$

M is mass of sun, $M=1.989\times 10^{30}\ kg$

So,

$T_1^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 1.496\times 10^{11}\\\\T_1=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times1.496\times10^{11}}\\\\T_1=2\times 10^{-4}\ s$

(b) For Sun -Mercury,

$T_2^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.79 \times 10^{10}\ m\\\\T_2=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.79 \times 10^{10}}\ m\\\\T_2=1.31\times 10^{-4}\ s$

(c) For Sun-Pluto,

$T_3^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times 5.91 \times 10^{12}\\\\T_3=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.989\times10^{30}}\times 5.91 \times 10^{12}}\\\\T_3=1.32\times 10^{-3}\ s$

8 0

3 years ago

Help? its due in like 12 minutes lolzz

Gnom [1K]

Answer:

Question 1: the plates are moving toward one another.

Question 2: The Himalayan Mountains in India

Question 3: Because mountains are formed instead.

Explanation:

The paragraph explains that the plates continue to move closer to one another while forming multiple mountains.

The paragraph explains, " a well-known example of this is the formation of the Himalayan Mountains in India,"

The area of the Himalayan Mountains are better suited for the formation of mountains rather than volcanoes.

Have a nice day!! Good Luck!! Brainliest would be appreciated!!!

4 0

3 years ago

The friends take a few minutes to discuss their ideas. Which of Amy's comments with respect to a charged particle moving in a ma

Marizza181 [45]

Answer:

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.

Explanation:

The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law $F=qv\times B$

From it we can clearly see that the magnitude of the magnetic force exerted on the particle is proportional to the magnitude of the charge q and to the speed v of the particle, and that it is also perpendicular to the particle's velocity. This means that at each instant it moves perpendicularly to the force, so the work done by the magnetic force on the particle is zero.

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.

7 0

3 years ago

What is the magnitude of the resultant of a 7.0-N force acting vertically upward and a 5.0-N force acting horizontally.

irakobra [83]

Answer:

Resultant force = 8.6N

Explanation:

Using Pythogorus' theorem

${a}^{2} + {b}^{2} = {c }^{2} \\ {7}^{2} + {5}^{2} = {r}^{2} \\ 49 + 25 = {r}^{2} \\ 74 = {r}^{2} \\ \sqrt{74} = \sqrt{ {r }^{2} } \\ 8.6 = r \\$