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wel
3 years ago
15

A dog starts from rest to chase after a ball. If the dog reaches a maximum speed of 5.4 m/s in 8.2 seconds, what is the accelera

tion of the dog?
Physics
1 answer:
Lelechka [254]3 years ago
4 0

Answer:

0.66 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0

Final velocity (v) = 5.4 m/s

Time (t) = 8.2 secs

Acceleration (a) =....?

The acceleration of the dog can be obtained as follow:

v = u + at

5.4 = 0 + a × 8.2

5.4 = 8.2a

Divide both side 8.2

a = 5.4/8.2

a = 0.66 m/s²

Therefore, the acceleration of the dog is 0.66 m/s².

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3 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
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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}}

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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

6 0
3 years ago
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