Well I don't know !
Let's work it out.
The gravitational force between two objects is
F = G · M₁·M₂ / R² .
'G' is the 'universal gravitational constant'. We could look it up.
'M₁' is the mass of one object
'M₂' is the mass of the other object
'R' is the distance between their centers.
It looks complicated, but stay with me. We can do this !
We know all the numbers, so we can calculate the force.
'G' is 6.67 x 10⁻¹¹ newton·meter² / kg² (I looked it up. You're welcome.)
'M₁' is 15 kg
'M₂' is 15 kg
'R' is 0.25 meter.
Now it's time to pluggum in.
F = G · M₁·M₂ / R²
= (6.67 x 10⁻¹¹ newton·meter² / kg²) · (15 kg) · (15 kg) / (0.25 m²)
= (6.67 x 10⁻¹¹ · 15 · 15 / 0.0625) N·m²·kg·kg / kg²·m²
= 2.4 x 10⁻⁷ Newton .
That a force equivalent to about 0.00000086 of an ounce.
This is the answer to part-a.
Concerning the answer to part-b ...
Personally, I could not detect this force, no matter what kind of equipment
I had. But I am just a poor schlepper engineer, educated in the last Century,
living out my days on Brainly and getting my kicks from YouTube videos.
I am not pushing the box to the envelope, or thinking outside the cutting
edge ... whatever.
I am sure there are people ... I can't name them, because they keep a
low profile, they stay under the radar, they don't attract a lot of media
attention, their work is not as newsworthy as the Kardashians, and plus,
they seldom call me or write to me ... but I know in my bones that there
are people who have measured the speed of light to NINE significant figures,
aimed a spacecraft accurately enough to take close-up pix of Pluto ten years
later, and detected gravity waves from massive blobs that merged 13 billion
years ago, and I tell you that YES ! THESE guys could detect and measure
a force of 0.86 micro-ounce if they felt like it !
Answer:
The answer is E) 6.7m at an angle 63° north of east
Explanation:
We begin by writing out our given vectors A,B and C in their respective rectangular Cartesian coordinate form. which are given as A=8i, B=6j and C= -5j (negative because it points towards west which is towards the left which is taken as the negative direction).Now we simply add all three to obtain our final vector given by,

In the last step we just added the like components. Now we will have to calculate the magnitude and angle of V(with respect of the east direction), which is given by,

the angle alpha is always calculated with respect to the horizontal positive axis, and since both components of the vector are positive the vector lies in the first quadrant. Hence the 63° angle north of east is justified. The term north of east means the angle formed in moving from east towards the north that is from the positive horizontal axis to the positive vertical axis.
The spring constant is 66.7 N/m
Explanation:
First of all, we have to find the magnitude of the force acting on the spring. This is equal to the weight of the mass hanging on the spring, which is:

where:
m = 0.50 kg is the mass of the object
is the acceleration of gravity
Substituting,

Now we can use Hookes' law to find the constant of the spring:

where
F is the force applied
k is the spring constant
x is the stretching of the spring
Here we have:
F = 5 N
While the stretching is
x = 0.075 m
Therefore, ignoring the negative sign in the formula (which only tells us the direction), we find the spring constant:

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