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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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2 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

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If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h

nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

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put here value

Vy = $\sqrt{35^2- 33^2}$

Vy = 11.66 m/s

and

now we get height

H = $\frac{Vy^2}{2g}$ .............................2

put here value

H = $\frac{11.66^2}{2\times 9.8}$

H = 6.93 m

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2 years ago

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