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3 years ago

How to find the energy of an 8 pound weight with e=mc2

1 answer:

7 0

Change the 8 pounds to kilograms (divide it by 2.2). Then multiply the kg by the speed of light (300,000,000 m/sec) squared. You get a very big number. It's the number of joules of energy equivalent to 8 lbs of mass.

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Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh

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Answer:

You cant draw on brainly

Explanation:

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2 years ago

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The density of an object is dependent upon the object’s mass and ---

kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

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3 years ago

If the pendulum is brought on the moon where the gravitational acceleration is about g/6, approximately what will its period now

Andru [333]

Answer:

The new period will be √6 *T

Explanation:

period ,T=2π√(L/g) ................equation 1

where T is the period on earth

gravitational acceleration on the moon is g/6

T1 = 2π√[L/(g/6)]

T1=2π√(6L/g) ...............equation 2

divide equation 2 by 1

T1/T =2π√(6L/g)÷2π√(L/g)

T1/T =√(6L/L)

T1/T =√6

T1 = √6 *T

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4 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

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3 years ago

True or false? The way in which a star forms does not depend on its mass

Lunna [17]

true because they wont be able to comperes hard enough to set off little nuclear explosins

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3 years ago

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