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2 years ago

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

Physics

1 answer:

Elanso [62]2 years ago

6 0

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]

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

$x = 12040\ m$

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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timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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3 years ago

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Hunter-Best [27]

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2 years ago

The figure above shows 4forces 3N, 10N, 3√3N, and 6N acting on a particle P. The resultant of the four forces is.

Brilliant_brown [7]

As you gave no pic I took them on one lined

F_1=3N
F_2=10N
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F_4=6N

$\\ \sf\longmapsto F_{net}=F_1+F_2\dots$

$\\ \sf\longmapsto F_{net}=3+10+6+3\sqrt{3}$

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2 years ago

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. if the ball hits the ground 4.0 second

Viktor [21]

Answer:

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Explanation:

To obtain the height of the cliff;

We can use the Relation to obtain the final velocity, v

v = u + at

a = acceleration due to gravity = 9.8m/s²

v = 0 + (9.8*4)

v = 0 + 39.2

v = 39.2 m/s

To obtain the Height, S

v² = u² + 2aS

39.2^2 = 0 + 2(9.8)S

39.2^2 = 0 + 19.6S

1536.64 = 19.6S

S = 1536.64 / 19.6

S = 78.4 m

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2 years ago

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stiv31 [10]

The hint is wrong.

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For the cylinder in the question,

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   = ( π ) · (11.925 cm³)

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2 years ago