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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 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.
Y_Kistochka [10]

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

a=\frac{v^{2} }{r}, where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

a=\frac{3.5^{2} }{8.5} \\a=1.4

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

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7 0
3 years ago
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Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

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Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

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y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

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