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zimovet [89]
3 years ago
11

A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

f 3.24 km. The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance.

Physics
1 answer:
Scorpion4ik [409]3 years ago
6 0

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

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The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h. (a) If the train goes
faust18 [17]

Answer:

a)  r = 6122 m and b) v = 32.5 m / s

Explanation:

a) The train in the curve is subject to centripetal acceleration

         a = v2 / r

Where v is The speed and r the radius of the curve

They indicate that the maximum acceleration of the person is 0.060g,

        a = 0.060 g

        a = 0.060 9.8

        a = 0.588 m /s²

Let's calculate the radius

        v = 216 km / h (1000m / 1km) (1 h / 3600 s =

        v = 60 m / s

        r = v² / a

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        r = 6122 m

b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m

        v = √a r

        v = √( 0.588 1800)

        v = 32.5 m / s

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