The potential energy is 490 because
1•9.8•50=490 hope it helps!
To solve this problem we will apply the kinematic equations of linear movement. For this purpose we will begin to define the final speed of the body before hitting the street. The first equation will begin using the difference in velocities as a function of acceleration (gravity) and position. And the second will use the concept of acceleration, time and speed, to find the time variable.
PART A) Equation of motion is
Replacing,
The speed of rock before hitting the ground is 32.74m/s
PART B) Equation of motion
Therefore the time taken by the rock is 5.58s
Answer:
a) B = 10⁻¹ r
, b) B = 4 10⁻⁹ / r
, c) B=0
Explanation:
For this exercise let's use Ampere's law
∫ B. ds = μ₀ I
Where I is the current locked in the path. Let's take a closed path as a circle
ds = 2π dr
B 2π r = μ₀ I
B = μ₀ I / 2μ₀ r
Let's analyze several cases
a) r <Rw
Since the radius of the circumference is less than that of the wire, the current is less, let's use the concept of current density
j = I / A
For this case
j = I /π Rw² = I’/π r²
I’= I r² / Rw²
The magnetic field is
B = (μ₀/ 2π) r²/Rw² 1 / r
B = (μ₀ / 2π) r / Rw²
calculate
B = 4π 10⁻⁷ /2π r / 0.002²
B = 10⁻¹ r
b) in field between Rw <r <Rs
In this case the current enclosed in the total current
I = 0.02 A
B = μ₀/ 2π I / r
B = 4π 10⁻⁷ / 2π 0.02 / r
B = 4 10⁻⁹ / r
c) the field outside the coaxial Rs <r
In this case the waxed current is zero, so
B = 0
Answer:
A)- SHORT WAVELENGTH AND LOW AMPLITUDE.
Explanation:
high pitch sound waves possess low wavelength and
quiet sound means that amplitue of the sound wave is low.
therefore option A matches the answer most.
Answer:
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