The displacement of Edward in the westerly direction is determined as 338.32 km.
<h3>What is displacement of Edward?</h3>
The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.
The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰
The displacement is the side of the triangle facing 150⁰ = R
R² = a² + b² - 2abcosR
R² = 150² + 200² - (2x 150 x 200)xcos(150)
R² = 62,500 - (-51,961.52)
R² = 114,461.52
R = 338.32 km
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Answer:
Δt = 5.29 x 10⁻⁴ s = 0.529 ms
Explanation:
The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

where,
Δt = required time interval = ?
Δs = distance between ears = 18 cm = 0.18 m
v = speed of sound = 340 m/s
Therefore,

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
= 
So,
Rearranging the above equation to get Electric field, we will get:
E = 
Multiply and divide by
E =
x 
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x
) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
= 
Rearranging the above equation for E:
E = 
E =
+ 
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E =
+ 
Replacing the above equations to get E:
E = (σ1 x
) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x
= - σ2 x 
Answer: Asking questions
Explanation:
Scientists always start with their question before
Observing anything
Answer: energy
Explanation: So the input and output of the power grid system is energy.