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skelet666 [1.2K]
4 years ago
7

While moving horizontally at 3.5 x 103 m/s at an altitude of 2.0 x 104 m, a ballistic missile explodes and breaks into two fragm

ents which fall freely. The smaller piece, which has 35% of the missile's total mass, falls straight down after the explosion and lands at point A, the place on the ground directly below the point of the explosion. How far from point A does the other fragment land?
Physics
1 answer:
Nastasia [14]4 years ago
4 0

Answer:

3, 43,968.88 m

Explanation:

Let mass of the missile be m

velocity = 3500 m/s

smaller part will have zero horizontal velocity and larger part will have velocity  v in horizontal direction

Applying conservation of momentum

m x 3500 = .35m x 0 + .65m x v

v = 5384.61 m / s

Height of  explosion

h = 20000 m

time to fall be t

for vertical fall u = 0 , g = 9.8

20000 = 1/2 x 9.8 t²

= 63.88 s

Horizontal range

time of fall x horizontal velocity

= 63.88 x 5384.61

= 3, 43,968.88 m

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Why is it important to understand if an organization is based on a centralized or decentralized organizational structure?
FrozenT [24]
Mostly because you need to understand the chain of command.
In a centralized organization, the decision making process fall to the hands of a very few people (in some case, it even can fall to one person) , which mean all level of managers can't move freely without the decision maker approval.
In a Decentralized organization, all different managers from different divisions have the authority to make decision
7 0
3 years ago
miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and
Tamiku [17]

Answer:

12 m/s

Explanation:

From the question,

Applying the law of conservation of momentum,

total momentum before collision = Total momentum after collision

mu+Mu' = mv+Mv'........................... Equation 1

Where m = mass of the bullet, u = initial velocity of the bullet, M = combined mass of the gun and the puck, u' = initial velocity of the gun and the puck, v = final velocity of the bullet, v' = final velocity of the gun and the puck

make v the subeject of the equation

v = [(mu+Mu')-Mv']/m................. Equation 2

Given: m = 5.00 g = 0.005  kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s

Substitute these values into equation 2

v = [0-(0.12×0.5)]/0.005

v = -0.06/0.005

v = -12 m/s

The negative sign  can be ignored since we are looking for the speed, which has only magnitude.

Hence the speed of the bullet is 12 m/s

5 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
What is the weight of a person with a massa of 80kg
Archy [21]
784 Newtons or 176.37 lbs
5 0
3 years ago
The starter motor of a car engine draws an electric current of 16.0 A from the battery. The copper wire to the motor is 5.00 mm
Lemur [1.5K]

Answer:

The electric charge passes is 15.2 C.

Explanation:

The electric current is drawn from a battery = 16 A

The diameter of copper wire is = 5 mm

The length of copper wire is = 1.52 m

Running time of starter motor before the start of car engine = 0.95 s

Therefore, the current 16 A passes through the copper wire for 0.95 seconds.

The calculation of electric charge passes:

Charge = i \times t \\

= 16 \times 0.95 \\

= 15.2 C

6 0
3 years ago
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