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Mrrafil [7]
3 years ago
5

A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative

velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship? Answer in MHz with no decimal places.
A) 55 MHz
B) 60 MHz
C) 66 MHz
D) 71 MHz
E) 76 MHz
Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

To first order  f' = f (1 - u/c) where observer and source are separating

f' = (1 + .4) = 70 MHz   so (d) would be correct

The next term would be 1/2 (u/c)^2 = 1/2 * ,4^2 = .08 and the correction would be .08 * 50 = 4 Mhz

Most approximations would use (d).

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A lorry of mass 4000 kg is travelling at a speed of 4 m/s.
bulgar [2K]

Answer:

KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

KE = .5*4000kg*(4m/s)²

KE = 32,000J

32,000J = .5*1000kg*v²

v² = 64

v = 8m/s

7 0
3 years ago
Police radar equipment is used to detect the speed of objects. In one trial, the radar equipment records a stationary tree as tr
Brut [27]

Answer:

reliability

accuracy

Explanation:

If a reading of a measurement is consistently the same then the measurement is reliable.

If a reading of measurement is close the actual value of the measurement then the reading is accurate.

Here, a stationary tree shows reading 6 mph once and 0 mph another instant. So, neither the reading of a measurement is consistent not the reading of measurement is close the actual value.

Hence, the radar has problems in its reliability and accuracy

8 0
3 years ago
What is the numerical value of the slope of the line (without units)?<br>0.083<br>24<br>8.3<br>12
tankabanditka [31]
As close as I can read it, it appears to be

                 1/12  gram/second

           (0.08333... gm/sec)
4 0
3 years ago
If a girl is standing still and holding a box, is she doing any work? Why or why not?
sukhopar [10]
I beleive she isnt doing any work due to holding the box motionless, you must be exerting a force in the direction of the box motion. If she is just standing there holding the box their isn't no work becuase no distance has been covered. work = force = distance.


7 0
3 years ago
Read 2 more answers
Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/
madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

8 0
3 years ago
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