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3 years ago

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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4 years ago

5. Suppose a skydiver (mass = 75kg) is falling toward the Earth. When the skydiver is 100m above the Earth he is moving at 60m/s

Anna71 [15]

1) Gravitational potential energy: 73 500 J

2) Kinetic energy: 135 000 J

3) Mechanical energy: 208 500 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the Earth's gravitational field, and it is given by

$U=mgh$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object relative to the ground

For the skydiver in this problem,

m = 75 kg is his mass

h = 100 m is his height above the Earth

Substituting, we find

$U=(75)(9.8)(100)=73500 J$

The kinetic energy of a body is the energy due to its motion, and it is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the skydiver in this problem,

m = 75 kg is his mass

v = 60 m/s is his speed

Substituting, we find the kinetic energy

$K=\frac{1}{2}(75)(60)^2=135000 J$

The mechanical energy of an object is the sum of the gravitational potential energy and the kinetic energy of the object:

$E=U+K$

where

U is the gravitational potential energy

K is the kinetic energy

For the skydiver in this problem,

U = 73,500 J is the gravitational potential energy

K = 135,000 J is the kinetic energy

Substituting, we find

$E=73,500 + 135,000 = 208,500 J$

Learn more about potential energy and kinetic energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

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4 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

4 years ago

What is the radioactive element used in nuclear war heads​

What is the radioactive element used in nuclear war heads