Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
<h2>
Answer:</h2>
143μH
<h2>
Explanation:</h2>
The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
<em>From the question;</em>
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
<em>But;</em>
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
<em>Substitute these values into equation (i) as follows;</em>
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
Answer:
I just took the quiz and got 100% when choosing A.Conservation. Hope this helps:)
Answer:
Two scientists in a lab examining vials of urine they are analyzing for levels of excreted protein.
Explanation:
Voltage - V = IR : v= voltage i= current r= resistance
