A change in momentum is also called: a. Impact b. Imput c. Impulse d. Impole

How does the atmospheric temperature and size of inner planets compare with those of outer planets?

amid [387]

The inner planets are rocky and have diameters of less than 13,000 kilometers. The outer planets include Jupiter, Saturn, Uranus, and Neptune. The smaller, inner planets include Mercury, Venus, Earth, and Mars. Inner planet's atmosphere is thin. (Mercury has no atmosphere). Outer Planets: Outer planets' atmosphere is very thick. The four inner planets, Mercury, Venus, Earth, and Mars, are warmer than the outer gas giants. However, the temperature of the planets does not follow a linear path from the Sun.

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Please give Brainliest!

6 0

3 years ago

Two small identical conducting spheres are placed with their centers 0.41 m apart. One is given a charge of 12 ✕ 10−9 C, the oth

nataly862011 [7]

(a) $-1.48\cdot 10^{-5}N$

The electrostatic force exerted between the two sphere is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges on the two spheres

r is the separation between the centres of the two spheres

In this problem,

$q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m$

Substituting these values into the equation, we find the force

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N$

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

(b) $+1.62\cdot 10^{-6}N$

The total net charge over the two sphere is:

$Q=q_1 +q_2 = 12\cdot 10^{-9}C+(-23\cdot 10^{-9}C)=-11\cdot 10^{-9} C$

When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

$q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C$

So the electrostatic force between the two spheres will now be

$F=k\frac{q^2}{r^2}$

And substituting numbers, we find

$F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N$

and the positive sign means the force is repulsive, since the two spheres have same sign charges.

7 0

3 years ago

This is a small rocky body that orbits the sun, and when close to the sun exhibit a small tail of ice and gas.

Over [174]

This is a comet. It is also called a dirty snowballs.

3 0

3 years ago

To measure the height of a building without a ruler or tape measure, an engineer drops a rock off the top of the building and fi

serg [7]

The relevant equation we can use in this problem is:

h = v0 t + 0.5 g t^2

where h is height, v0 is initial velocity, t is time, g is gravity

Since it was stated that the rock was drop, so it was free fall and v0 = 0, therefore:

h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2

h = 117.77 m

6 0

3 years ago

If an object accelerates from rest, with a constant acceleration of 4.4 m/s2, what will its velocity be after 28s?

Norma-Jean [14]

Answer:

123.2 m/s after 28s

Explanation:

Vi= 0 m/s

a= 4.4 m/s^2

t=28s

Vf after 28s

To find Vf use your kinematics formula Vf=Vi+at

Vi is Zero so it gets removed and the equation becomes

Vf=at

Simply Plug and Solve

Vf= 4.4(28)

Vf=123.2 m/s after 28s

6 0

3 years ago