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Blababa [14]
3 years ago
12

8. A sprinter on a school track team is running north at a velocity of 6.0 m/s. After 5.0 s, she

Physics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

acc. = 4-(-6) /5= 10/5=2 m/s^2

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2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?
Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

v=\frac{d}{t}

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

v=\frac{15 km}{2.0 h}=7.5 km/h

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

v=\frac{15,000 m}{7200 s}=2.1 m/s

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A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge
Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

\tau=RC

where

R=50,000 \Omega is the resistance

C=2.0\mu F=2.0\cdot 10^{-6}F is the capacitance

Substituting,

\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s

The charge on a charging capacitor is given by

Q(t)=Q_0 (1-e^{-t/\tau} ) (1)

where

Q_0 is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Q(t)=0.90 Q_0

Substituting this into eq.(1) we find

0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s

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