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Dmitrij [34]
3 years ago
12

Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave

ls at 8 m/s. How long does it take for the two cars to meet
Physics
1 answer:
mylen [45]3 years ago
7 0

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

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The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

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A pitcher throws a 0.145-kg baseball at a velocity of 30.0m/s. How
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Answer: 65.25 J

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3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

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A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

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