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SIZIF [17.4K]
3 years ago
14

Convection currents within Earth's core and mantle transfer large amounts of heat when

Physics
1 answer:
liubo4ka [24]3 years ago
8 0
<span>c. hotter, less dense rock sinks and cooler, more dense rock rises.</span>
You might be interested in
A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.
SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0
4 years ago
How many electrons can the first shell hold
Naily [24]
First shell hold up to 2 electrons.
8 0
4 years ago
Plzz helpp mee!!!! plzzz
Gnesinka [82]

If you add 2 miles from west then 2 miles east then it would 4 miles all together.

7 0
3 years ago
Two cars are traveling around identical circular racetracks. Car A travels at a constant speed of 20 m/s. Car B starts at rest a
Tomtit [17]

Answer:

b. it has the same centripetal acceleration as car A.

Explanation:

According to the question, the data provided is as follows

Constant speed of car A = 20 m/s

Constant tangential acceleration until its speed is 40 m/s

Based on the above information, the true statement is the same centripetal acceleration as car A because

As we know that

Centripetal acceleration is

= \frac{V^2}{r}

where,

V^2 = velocity

r = radius of the path

Now if both car A and car B moving in the same or identical circular path having the same velocity so in this case there is the same centripetal acceleration for that particular time

hence, the second option is correct

3 0
4 years ago
A grapefruit falls from a tree and hits the ground 0.72 s later.
xxTIMURxx [149]

Answer:

<em>The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.8 m/s^2.

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

\displaystyle y=\frac{gt^2}{2}

Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:

\displaystyle y=\frac{9.8\cdot 0.72^2}{2}

y = 2.54 m

The final speed is computed below:

vf=9.8\cdot 0.72

vf = 7.06 m/s

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

6 0
3 years ago
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