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Allisa [31]
3 years ago
5

Why is plate tectonics a widely accepted theory?

Physics
1 answer:
MariettaO [177]3 years ago
6 0

Answer: The theory of Plate Tectonics is now widely accepted because there is sufficient proof to support it, and it is an important aspect of geology, oceanography, geophysics and even paleontology.

Explanation: In places where a plate faced resistance to its movement, it would fold upward and create mountains. Hope this helped! :)

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Moustapha jones drives east at 100km/hr for 3 hours then back west at 80km/hr for 1.5 hours. which pair of answers gives his ave
olya-2409 [2.1K]

The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

The total distance cover in east direction is=100*3=300 km

The total distance cover in the west direction=80*1.5=120 km

The total distance covered is =300+120=420 km

And Total displacement of the car is =300-120=180 km

As we know that the average speed is given as

Avg Speed =Total Distance / Total time

=420/4.5=93.33 km/hr

As we know that the average velocity is given as

Avg Speed =Total Displacement/ Total time

=180/4.5=40 km/hr

Therefore, The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

7 0
3 years ago
Which group has the Same electron shells? A Titanium and Gold B Selenium and Tin C Calcium and Magnesium D Lithium and Boron
Anna11 [10]

im not 100 percent sure but i think its b


6 0
3 years ago
Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:
miss Akunina [59]

Answer:

good morning you are you still

8 0
3 years ago
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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