In a stadium, fans stand up and sit down to produce a wave across the stadium. This type of wave where the material travels perp
2 answers:
You might be interested in
To solve this problem we will apply the relationship between Newton's second law and Hooke's law, with which we will define the balance of the system. From the only unknown in that equation that will be the constant of the spring, we will proceed to find the period of vibration of the car.
We know from Hooke's law that the force in a spring is defined as
Here k is the spring constant and x the displacement
While by Newton's second law we have that the Weight can be defined as
Here m is the mass and g the gravity acceleration.
The total weight would be
Each spring takes a quarter of the weight, then
Since the system is in equilibrium the force produced by the weight in each spring must be equivalent to the force of the spring, that is to say
The period of a spring-mass system is given as
The total mass is equivalent as the sum of all the weights, then replacing we have that the Period is
Therefore the period of vibration of the car as it comes to rest after the four get in is 0.9635s
Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3
Answer:
R₁ = 0.126 m
Explanation:
Let's use the definition of intensity which is the power per unit area
I = P / A
the generated power is constant
P = I A
power is
P = E / t
if we perform the calculations for a given time, the wave energy is
E = q V
we substitute
P =
we can write this equation for two points, point 1 the antenna and point 2 the receiver
V₁A₁ = V₂A₂
A₁ =
A₁ = 0.1 10⁻³ 5 10⁻⁴ /V₁
A₁ = 5 10⁻⁸ /V₁
In general, the electric field on the antenna is very small on the order of micro volts, suppose V₁ = 1 10⁻⁶ V
let's calculate
A₁ = 5 10⁻⁸ / 1 10⁻⁶
A₁ = 5 10⁻² m²
the area of a circle is
A = π r²
we substitute
π R1₁²= 5 10⁻²
R₁ =
R₁ = 0.126 m