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Ira Lisetskai [31]

2 years ago

13

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 30 mph. at the same tim

e, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. the officer's radar gun indicates 15 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 15 mph). what is the speed of the other car?

2 answers:

IgorLugansk [536]2 years ago

6 0

Answer:

$v = 51.2 mph$

Explanation:

Velocity of police car

$v_p = 30 mph$ towards North

velocity of the car with respect to police car

$v_{cp} = 15 mph$

Speed of the other car is given with respect to police car so let say the speed of other car is "v"

Now the component of this speed along the line joining two cars is

$v_1 = v cos45$

similarly speed of police car along the line joining two cars is given as

$v_2 = v sin45$

now we can find the relative speed

$v_{12} = v_1 - v_2$

$15 mph = vcos45 - 30cos45$

$v = 51.2 mph$

aleksandr82 [10.1K]2 years ago

3 0

In triangle ABC , using Pythagorean theorem

BC = sqrt(AB² + AC²)

r = sqrt(y² + x²) eq-1

taking derivative both side relative to "t"

dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)

15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)

v₂ = 51.2 m/s

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0

2 years ago

An engineer wants to design an oval racetrack such that 3.20 × 10 3 lb racecars can round the exactly 1000 ft radius turns at 10

Reptile [31]

Answer:

The banking angle necessary for the race cars is 34.84°

Explanation:

For normal reaction the expression is:

$\\Nsin\theta = \frac{mv^{2} }{R} =Fc\\tan\theta =\frac{v^{2} }{Rg} \\\theta =tan^{-1} (\frac{v^{2} }{Rg} )\\\theta =tan^{-1} (\frac{(102*0.447)^{2} }{1000*0.3048*9.8} )=34.84$

4 0

1 year ago

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

yawa3891 [41]

Answer:

Isothermal : P2 = ( P1V1 / V2 ) , work-done $pdv = nRT * In( \frac{V2}{v1} )$

Adiabatic : : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$ , work-done =

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

$pdv = nRT * In( \frac{V2}{v1} )$

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

$P1V1^y = P2V2^y$

where y = 5/3

hence : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$

Work-done

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Where $T2 = T1V1^(2/3)/V2^(2/3)$

3 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

A 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800m/s when it makes a head-o

e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

$m_1$ = Mass of first glider = 0.3 kg

$m_2$ = Mass of second glider = 0.15 kg

$u_1$ = Initial Velocity of first glider = 0.8 m/s

$u_2$ = Initial Velocity of second glider = 0 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s$

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s$

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

$K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J$

Final kinetic energy of first glider is 0.010935 J

$K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J$

Final kinetic energy of second glider is 0.0858675 J

6 0

2 years ago