Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R = ![\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]](https://tex.z-dn.net/?f=%5Crho%20%28x%29%20%3D%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%20%5C%20m%7D%29%5E2%5D)
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = 
![dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}](https://tex.z-dn.net/?f=dR%20%3D%20%5Cfrac%7B%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%7D%29%5E2%5D%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%2810%5E%7B-3%7D%29%5E2%7D)
Taking the integral of both sides;we have:
![\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5ER_0%20%20dR%20%3D%20%5Cint%5Climits%5E2_0%203.185%20%5C%20%5B1%2Bx%5E2%5D%20%5C%20dx)
![R = 3.185 [x + \frac {x^3}{3}}]^2__0](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5Bx%20%2B%20%5Cfrac%20%7Bx%5E3%7D%7B3%7D%7D%5D%5E2__0)
![R = 3.185 [2 + \frac {2^3}{3}}]](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5B2%20%2B%20%5Cfrac%20%7B2%5E3%7D%7B3%7D%7D%5D)
R = 14.863 Ω
Since V = IR
I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
Answer:
30m/s
Explanation:
From law of motion equation
Vf= Vi + at
Where Vf= final velocity
Vi= initial velocity=0(the car started at rest)
a= acceleration= 3m/s2
t= time= 10s
Then substitute into the equation to get the final velocity.
Vf= 0+(10×3)
Vf= 30m/s
Hence, the car's final velocity is 30m/s
Nuclear decay formula is N(t)=N₀*2^-(t/T), where N(t) is the amount of nuclear material in some moment t, N₀ is the original amount of nuclear material, t is time and T is the half life of the material, in this case carbon 14. In our case N(t)=12.5% of N₀ or N(t)=0.125*N₀, T=5730 years and we need to solve for t:
0.125*N₀=N₀*2^-(t/T), N₀ cancels out and we get:
0.125=2^-(t/T),
ln(0.125)=ln(2^-(t/T))
ln(0.125)=-(t/T)*ln(2), we divide by ln(2),
ln(0.125)/ln(2)=-t/T, multiply by T,
{ln(0.125)/ln(2)}*T=-t, divide by (-1) and plug in T=5730 years,
{ln(0.125)/[-ln(2)]}*5730=t
t=3*5730=17190 years.
The bone is t= 17190 years old.
Source localization in ocean acoustics is posed as a machine learning problem in which data-driven methods learn source ranges directly from observed acoustic data: True.
<h3>What is machine learning?</h3>
Machine learning (ML) is also known as artificial intelligence (AI) and it can be defined as a subfield in computer science which typically focuses on the use of computer algorithms, data-driven techniques (methods) and technologies to develop a smart computer-controlled robot that has the ability to automatically perform and manage tasks that are exclusively meant for humans or solved by using human intelligence.
In Machine learning (ML), data-driven techniques (methods) are used to learn source ranges directly from observed acoustic data in a bid to proffer solutions to source localization in ocean acoustics.
In conclusion, a normalized sample covariance matrix (SCM) is constructed and used as the input, especially after pre-processing the pressure that's received by a vertical linear array in Machine learning (ML).
Read more on machine learning here: brainly.com/question/25523571
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