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3 years ago

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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?

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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Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

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PSYCHO15rus [73]

Answer:

Because the hiker walked directly west and then directly north the two legs of the hike forms a right triangle. Therefore we can use the Pythagorean Theorem to solve this problem.

c=5

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Two steel guitar strings have the same length. String A has a diameter of 0.513 mm and is under 403 N of tension. String B has a

Mnenie [13.5K]

Answer:

$\frac{v_{A}}{v_{B}} = 1.785$

Explanation:

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$T_{B}$ = Tension force in string B = 800 N

$d_{A}$ = diameter of string A = 0.513 mm

$d_{B}$ = diameter of string B = 1.29 mm

$v_{A}$ = wave speed of string A

$v_{B}$ = wave speed of string B

Ratio of the wave speeds is given as

$\frac{v_{A}}{v_{B}} = \sqrt{\frac{T_{A}}{T_{B}}} \left ( \frac{d_{B}}{d_{A}} \right )$

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Answer:

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