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jenyasd209 [6]
3 years ago
10

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

Physics
1 answer:
Elanso [62]3 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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What is the mass of an object that weighs 686N on Earth?
Oxana [17]

Answer:

70 kg is the mass of the object

Explanation:

This question can be solved with this simple formula:

Weight force = mass . gravity

686 N = mass . 9.8 m/s²

686 N /  9.8 m/s² = mass → 70 kg

Note → 1N = 1 kg . m / s²

8 0
3 years ago
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0
3 years ago
Caffeine concentration is 1.99 mg/oz how many cans would be leathal if 10g was leathal and there where 12oz in a can
katovenus [111]

The number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

<h3>How to convert mass?</h3>

According to this question, caffeine concentration is 1.99 mg/oz.

1.99 milligrams can be converted to grams as follows:

1.99milligrams ÷ 1000 = 0.00199grams

This means that 0.00199grams per oz is the caffeine concentration.

If there were 12 oz in a can, then, 0.00199grams × 12 = 0.02388 grams in 1 can.

This means that if 10grams is considered lethal, 10grams ÷ 0.02388 grams = 419 cans would be lethal for consumption.

Therefore, the number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

Learn more about conversion factor at: brainly.com/question/14479308

#SPJ1

5 0
1 year ago
A firework is launched with a force of 700 N and a momentum of 200 kg-m/s. How much time before is explodes?
8_murik_8 [283]

Answer:

t = 0.28 seconds

Explanation:

Given that,

Force acting on a firework, F = 700 N

The momentum of the firework, p =200 kg-m/s

We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,

F=\dfrac{P}{t}\\\\t=\dfrac{P}{F}\\\\t=\dfrac{200}{700}\\t=0.28\ s

So, the required time is equal to 0.28 seconds.

7 0
3 years ago
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe
gladu [14]

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

4 0
3 years ago
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