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Mademuasel [1]
3 years ago
14

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

her bomb is tested, and 3.6 grams of matter are converted into energy. How many joules of energy are released by the explosion?
Physics
1 answer:
Softa [21]3 years ago
3 0

Answer:

1.08 * 10^{14} J

Explanation:

Energy and mass are related by the famous equation developed by Albert Einstein:

E = mc^2

where m = mass and c = speed of light

This equation explains that an object with very small mass can produce a large amount of energy in reactions such as a nuclear reaction.

Hence, the energy produced by the explosion of a Plutonium bomb containing 3.6 grams of matter is:

E = 3.6 * 10^{-3} * (3 * 10^8)^2

E = 1.08 * 10^{14} J

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Bigger change in velocity because the object is lighter than the object with more mass so it would move further (sorry it’s not a great explanation)
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A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8
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Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s

(a).

F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})

(b). T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s

(c). v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J

3 0
3 years ago
Read 2 more answers
A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to
leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s

So, the combined velocity is 8.61 m/s.

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3 years ago
What traits did Sir William Gilbert have that made him a good scientist?
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Gave him good advice to others
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PLEASE HELP WHICH STATEMENTS ARE CORRECT DO NOT GUESS
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THe first one and the third one!!
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