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bija089 [108]
4 years ago
6

As a defensive measure, forces afloat would typically operate in highly dispersed formations, sometimes covering an area of more

than 40 square miles.A. True.B. False
Physics
1 answer:
hoa [83]4 years ago
7 0

Answer:

B - False

Explanation: a dispersed formation means spreading out, more like an attacking or advancing position. This doesn't depict a defensive measure.

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How much force to move 18 kg at a rate of 3m/s​
Stolb23 [73]

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

4 0
3 years ago
A pitcher throws a softball toward home plate. When the ball hits the catcher’s mitt, its horizontal velocity is 32 meters/secon
n200080 [17]
To calculate the force of impact F, first lets calculate the acceleration a of the ball: 

a=v/t where v is the velocity of the ball and t is time

a=32/0.8=40 m/s²

To get the force F we need the Newtons second law:

F=m*a where m is the mass of the ball and a is the acceleration.

F=m*a= 0.2*40 = 8 N

So the impact force is F= 8 N.
5 0
3 years ago
Read 2 more answers
2. Why is natural selection a mechanism for biological change?
Katena32 [7]

Answer:

Natural selection is a simple mechanism that causes populations of living things to change over time.  Organisms that are more adapted to their environment are more likely to survive and pass on the genes that aided their success.

4 0
2 years ago
Three objects each with a mass of 10.0 kg are
Vera_Pavlovna [14]

Answer: F net=0

Explanation:

Its in the picture.

7 0
3 years ago
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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)

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

6 0
3 years ago
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