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3 years ago

What is a resultant vectors direction?

2 answers:

8 0

That completely depends on all sizes and all directions of all of the vectors that combined to produce the resultant one.

vovangra [49]3 years ago

5 0

It depends. I do not know if you are adding vectors in 1D, 2D or even 3D.
In 1D: its direction is along the x axis if its positive, otherwise it is in the opposite direction
In 2D: its direction is along the x or y axis if its component is positive, otherwise it is in the opposite direction
In 3D: its direction is along the x, y or z axis if its component is positive, otherwise it is in the opposite direction

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Two rooms are in contact. One is at 15°C and the other at 21°C. Choose the best statement concerning the heat flow.

amid [387]

A is the answer to your question

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3 years ago

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Earth's tides are primarily caused by _____.

Fiesta28 [93]

B.) The moon

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4 years ago

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You walk 53 m to the north, then you turn 60° to your right and walk another 45 m. Determine the direction of your displacement

Anton [14]

Answer:

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3 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

The formula d = 1.1 t 2 + t + 1 expresses a car's distance (in feet to the north of an intersection, d , in terms of the number

Phantasy [73]

d(t) = 1.1t² + t + 1

The constant speed required to cover the same distance between t = 3 to t = 5 is the same as the average speed over that same time interval. It is given by:

v = Δx/Δt

v = average speed, Δx = change in distance, Δt = elapsed time

Given values:

Δx = d(5) - d(3) = 19.6ft

Δt = 5s - 3s = 2s

Plug in and solve for v:

v = 19.6/2

v = 9.8ft/s

5 0

4 years ago

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