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-BARSIC- [3]
3 years ago
15

A bomb is dropped from a bomber traveling at the speed of 120 km / h, destroying a military objective located at a distance of 2

000 m. How high was the plane traveling?
Physics
1 answer:
schepotkina [342]3 years ago
7 0

Answer:

18 km

Explanation:

Convert km/h to m/s:

120 km/h × (1000 m/km) × (1 h / 3600 s) = 33.3 m/s

The time it takes the bomb to travel the 2000 meters is:

2000 m / (33.3 m/s) = 60 s

So it takes 60 seconds for the bomb to fall.  The distance it fell is therefore:

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) (60 s) + ½ (10 m/s²) (60 s)²

Δy = 18,000 m

Δy = 18 km

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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?
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4 0
3 years ago
Read 2 more answers
Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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