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Pachacha [2.7K]
3 years ago
15

Why do u think you usually can't feel or notice the movement of the continents

Physics
1 answer:
luda_lava [24]3 years ago
7 0
You can usaly not feal it because the earth is moving and the plates aren't having any major change to the movement.
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Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0 ​
Marina CMI [18]
Same directions
Perpendicular
Opposite directions

5 0
3 years ago
A 90-kg astronaut is stranded in space at a point 6.0 m from his spaceship, and he needs to get back in 4.0 min to control the s
Stella [2.4K]

Momentum = 0.5 * 4 = 2 
to conclude the man’s velocity after he throws the piece of equipment, divide this number by the man’s mass. 

v = 2/90 

This is about 0.0222 m/s. To know if he can move 6 meters at velocity in 4minutes, use the following equation. 

d = v * t, t = 4 * 60 = 240 s 
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This is ⅔ of a meter from the spaceship. To know the velocity that he must have to move 6 meter, use the same equation. 

6 = v * 240 
v = 6/240 
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To know the velocity of the package, divide this number by the mass of the package. 
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8 0
3 years ago
Mixing salt in water is an example of physical change
galina1969 [7]

EXPLAIN MORE!!! i need more detail if i can help you

6 0
3 years ago
A woman does 236 J of work
Ratling [72]

Answer:

The woman's force was directed 59.22⁰ to the horizontal.

Explanation:

Given;

work done by the woman, W = 236 J

distance through the load was moved, d = 24.4 m

applied force, F = 18.9 N

inclination of the force, = θ

The work done by the woman is calculated as;

W = Fdcosθ

cos \ \theta = \frac{W}{Fd} \\\\cos \ \theta = \frac{236 }{18.9 \times 24.4} \\\\cos \ \theta =  0.5118\\\\\theta = cos^{-1} ( 0.5118)\\\\\theta =59.22^0

Therefore, the woman's force was directed 59.22⁰ to the horizontal.

8 0
3 years ago
a particle that has a mass of 2.5 kg is moving in the positive X-direction with a constant velocity of 1.6m/s. Suddenly a consta
malfutka [58]
The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.
X = x + vt + (1/2)at²
10 = 0 + (1.6)t + (1/2)(0)t²
10/1.6 = t
t = 6.25s

So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.
Start with Newton's second law.
F = ma
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a = 3.2m/s²

Now, use kinematics again.
Y = y + vt + (1/2)at²
Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²
Y = <u>62.5m</u>
7 0
3 years ago
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