Answer:
the center of mass is 7.07 cm apart from the bend
Explanation:
the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is
x₁ = L/2 = 20 cm /2 = 10 cm
when the wire is bent in a right angle the coordinates of the new centre of mass will be
x₂ = L₂/2
y₂= L₂/2
where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2
x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm
y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm
x₂=y₂=X
locating the bend in the origin (0,0) the distance to the centre of mass is
d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm
d = 7.07 cm
The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
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NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
</span><span>
KOH
- it is a strong base (pH=14)
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↓
He needs to use HF (Hydrogen fluoride) to decrease the pH.
We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:

Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!