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amm1812
2 years ago
5

Electrically charged particles are found primarily in

Physics
1 answer:
lesantik [10]2 years ago
8 0

Answer:

the ionosphere

.......................................................

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The arrows on the diagram show the ocean floor spreading from the Redge what three kinds of evidence scientists have found suppo
Blizzard [7]
Edit: You do mean Ridge?

Rocks near Mid-Ocean Ridge are younger than rocks near the trenches.

Seismic data shows oceanic crust is sinking into the mantle at the trenches.

Matching bands of magnetic rock are found on either side of the Ridge. Earth's magnetic fields change these bands over time.
5 0
2 years ago
What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 n
Hoochie [10]

Force on electron due to electric field is given by

F = eE

F = 1.6 * 10^{-19}* 6100

F = 9.76 * 10^{-16} N

now the acceleration is given by

a = \frac{F}{m}

a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}

a = 1.07 * 10^{15} m/s^2

so above is the magnitude of acceleration and its direction is opposite to field as electron is negatively charged so direction is towards SOUTH

4 0
3 years ago
Why a paperclip does NOT float?
Nataliya [291]
Because of Surface tension
3 0
2 years ago
Read 2 more answers
Ceres has an orbital semi-major axis = 2.768 AU. What is Ceres’ orbital period?
Doss [256]

Answer: 4.6 years

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a (size) of its orbit”: </em>

<em />

T^{2}\propto a^{3} (1)  

However, if T is measured in Earth years, and a is measured in astronomical units (unit equivalent to the distance between the Sun and the Earth), equation (1) becomes:  

T^{2}=a^{3} (2)  

Knowing a=2.768 AU and isolating T from (2):  

T=\sqrt{a^{3}} (3)  

T=\sqrt{(2.768 AU)^{3}} (4)  

Finally:  

T=4.6 years This is Ceres' orbital period

6 0
3 years ago
A person walks in the following pattern: 2.9 km north, then 2.8 km west, and finally 4.4 km south. (a) How far and (b) at what a
vfiekz [6]

Answer:

(a) Magnitude =3.18 km

(a) Angle =28.2°

Explanation:

(a) To find magnitude of this vector recognize

R_{x}=-2.8 km\\R_{y}=-1.5km

Use Pythagorean theorem

R=\sqrt{(R_{x})^{2}+(R_{y})^{2}  }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2}  }\\ R=3.18km

(b)To find the angle use the trigonometric property

tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}

5 0
3 years ago
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