In order to measure the size of the electrical current flowing in the circuit,
the current must pass through the meter.
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
Answer:
Tension = 0.012 N
Explanation:
If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,
Tension = Weight
Tension = mg
where,
m = mass of spider = 1.27 g = 0.00127 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
Tension = (0.00127 kg)(9.8 m/s²)
<u>Tension = 0.012 N</u>
Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :
E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
