The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.
<h3>What is spring force?</h3>
The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =200
K is the spring constant= 2500 N/m
x is the length by which spring got stretched =?
The stretch of the spring is found as;
Hence the length by which the spring got stretched will be 0.08 m.
To learn more about the spring force refer to the link;
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Answer:
his movement is proportional to the intensity of the earthquake,
Explanation:
An earthquake is a record of the intensity of an earthquake as a function of time.
Where the intensity is plotted on the y-axis, which corresponds to the vertical movement of the detector, this movement is proportional to the intensity of the earthquake, therefore the intensity increases the amplitude of the oscillation increases.
And the in x corresponds to time
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
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
