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harkovskaia [24]
3 years ago
8

Please select the word from the list that best fits the definition the name for continents in one landmass surrounded by giganti

c ocean
Physics
1 answer:
frez [133]3 years ago
8 0
Hello!

The name for continents in one landmass surrounded by gigantic ocean is Pangaea.

This name is originated from the Greek words Pan ("all") and Gaia ("Earth"). It describes the supercontinent that would have existed in the Cenozoic and Mesozoic periods by the joining of all continents that we know them today.

The gigantic ocean that surrounded it is called Panthalassa, from Greek Pan ("all") and thálassa ("sea"). 

Have a nice day!
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Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.
Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)=k q1q2/r^2

    k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m

So,    

        F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}

         F= 0.6 N

5 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam
Allushta [10]
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹ 
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
4 0
3 years ago
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Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

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now we know that

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M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

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5 0
3 years ago
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