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Novay_Z [31]
3 years ago
7

What is the radioactive element used in nuclear war heads​

Physics
2 answers:
diamong [38]3 years ago
7 0

Answer:

Plutonium, has formula Pu with molecular mass 94

Explanation:

.

liraira [26]3 years ago
3 0

The commonly used radio active elements in Nuclear warheads are

Plutonium:-

  • Symbol:-Pu
  • Atomic no=94
<h3>Uranium</h3>
  • Symbol=U
  • Atomic no=92

<h3>Thorium:-</h3>

  • Symbol=Th
  • Atomic no=92
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A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move up
Snezhnost [94]

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          t=\frac{200}{9.81}=20.38s

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

6 0
3 years ago
Explain why pumice floats and why ironwood sinks
almond37 [142]

Answer:

A quick way of describing density is to describe an object as heavy or light for its size. Pumice stone, unlike regular rock, does not sink in water because it has a low density. An ironwood branch is very dense and sinks in water.

Hope that helps. x

6 0
2 years ago
Read 2 more answers
What is the gravitational potential energy of a rock with the mass of 67 kg if it is sitting on top of a hill .35 kilometers hig
solong [7]
Not to sure but maybe 23.45
8 0
3 years ago
Which explains the downward force produced when air flows over the winglike spoiler on a race car?
professor190 [17]
Newton's principle
newton's 2nd law of motion where F=ma
3 0
3 years ago
Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5
lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0
3 years ago
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