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stellarik [79]
3 years ago
15

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

und lane moving with a speed of 11 m/s, and Car B is in the westbound lane moving with a speed of 31 m/s.
(a) What is the magnitude and direction of the velocity of car A as seen by the driver of car B? magnitude m/s direction
(b) What is the magnitude and direction of the velocity of car B as seen by the driver of car A? magnitude m/s direction
Physics
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
What is black body radiation? Explain in detail.
tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0
3 years ago
Which of the four spheres is this pictures/scene?
Zigmanuir [339]

I believe the correct answer is atmosphere (D).

7 0
3 years ago
Read 2 more answers
How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te
Tresset [83]

Answer:

<em>0.97c</em>

<em></em>

Explanation:

From the relativistic equation for length contraction, we have

l = l_{0}\sqrt{1 - \beta }

where

l is the final length of the object

l_{0} is the original length of the object before contraction

β = v^{2} /c^2

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

l/l_{0} = \sqrt{1 - \beta }

For the length to contract to one-fourth of the proper length, then

l/l_{0} = 1/4

substituting into the equation, we'll have

1/4 = \sqrt{1 - \beta }

substituting for β, we'll have

1/4 = \sqrt{1 - v^2/c^2 }

squaring both side of the equation, we'll have

1/16 = 1 - v^2/c^2

v^2/c^2 = 1 - 1/16

v^2/c^2 = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = <em>0.97c</em>

3 0
3 years ago
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