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:
truck 1 has the most velocity
Explanation:
Because it weights less which means it faster and yea
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
Learn more about Conservation of Linear Momentum here:
brainly.com/question/7538238
