The initial position of the object was found to be 134.09 m.
<u>Explanation:</u>
As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So
Displacement = Final position - Initial position.
As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be
Initial position of the object = Final position-Displacement
Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.
Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.
Answer:
a)
two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)
b)
both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Explanation:
Given that;
L = 0.26 m
k = 180 N/m
x = 0.039 m
a)
we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.
b)
Spring force F = kx
F = 180 × 0.039
F = 7.02 N
Now, Electrostatic force F = Keq²/r²
where r = L + x = ( 0.26 + 0.039 )
we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2
so from the equation; F = Keq²/r²
Fr² = Keq²
q = √ ( Fr² / Ke )
we substitute
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ (0.627595 / 9×10⁹)
q = √(6.97 × 10⁻¹¹)
q = 8.35 × 10⁻⁶ C
Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Answer:
to locate places on earth
Answer:
My mom always told me he was just there
Explanation:
Answer:
The x-component of 
is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
(1)
Where:
, 
, 
- External forces exerted on the ring, measured in newtons.
- Vector zero, measured in newtons.
If we know that ![\vec F_{1} = (70.711,70.711)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B1%7D%20%3D%20%2870.711%2C70.711%29%5C%2C%5BN%5D)
, ![\vec F_{2} = (-126.859, 46.173)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20F_%7B2%7D%20%3D%20%28-126.859%2C%2046.173%29%5C%2C%5BN%5D)
, 
and ![\vec O = (0,0)\,[N]](https://tex.z-dn.net/?f=%5Cvec%20O%20%3D%20%280%2C0%29%5C%2C%5BN%5D)
, then we construct the following system of linear equations:
(2)
(3)
The solution of this system is:
, 
The x-component of is 56.148 newtons.
