The kinetic theory states that the particles in matter are always in?

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

3 years ago

Read 2 more answers

a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q

Romashka-Z-Leto [24]

The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

5 0

1 year ago

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap

Verizon [17]

Answer:

The answer is " $1.144 \times 10^{-4} \ m$ ".

Explanation:

$w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\$

$=\frac{2.44 \times 633 \times 10^{-9}\times 4 }{0.054}\\\\=\frac{6178.08\times 10^{-9}}{0.054}\\\\=1.144 \times 10^{-4} \ m$

8 0

3 years ago

Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slid

galben [10]

Answer:

Explanation:

They have the same kinetic energy

5 0

3 years ago

A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and

marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

$\lambda = \frac{v}{f}$