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boyakko [2]
3 years ago
13

An iceberg of density 920kg/m^3 floats in seawater of density 1025kg/m^3 with volume of 10^6m^3.What is the total mass of the ic

eberg?
Physics
1 answer:
zhenek [66]3 years ago
5 0
Im pretty sure the answer is 1.95x10^9
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A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

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3 years ago
PLZZZ HELP 100 POINTS WILL MARK BRAINELST
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Explanation:

Crust...molten

a. Oceanic, iron

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a. Lithosphere

b. Asthenosphere

4. Core

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Couldn't answer all but wanted to help

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All but one statement applies to electromagnetic radiation
lina2011 [118]
<span>D) Electromagnetic radiation travels in the form of longitudinal waves.</span>
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3 years ago
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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3 years ago
Which action is mostly likely to make the team on the left win
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D. adding one student to the team on the left
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