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3 years ago

Scientist accept ideas of the past and do not try to revise or change them

Physics

1 answer:

Vikentia [17]3 years ago

4 0

Is this a question or a statement. This statement would be incorrect. <span />

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

Nastasia [14]

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.

7 0

3 years ago

An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops

Alina [70]

To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,

$PE = KE$

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

Here,

m = Mass

g = Gravity

h = Height

v = Velocity

Rearranging to find the velocity

$v = \sqrt{2gh}$

Replacing,

$v = \sqrt{2(9.8)(120)}$

$v = 48.49m/s$

Using the vector properties the magnitude of the velocity vector would be given by,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{45^2+48.42^2}$

$|V| = 66.2m/s$

Therefore the package is moving to 66.2m/s

3 0

3 years ago

Can someone help me with this question?

padilas [110]

Answer:

it is 5

Explanation:

5 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

3 years ago

A man pushes a lawn mower on a level lawn with a force of 250N. If 40% of this force is directed downward, how much work is done

Dvinal [7]

When considering work, we always take the force directed along the axis of motion (in this case, the horizontal axis). If 40% of the force is directed downward, then 60% of the force is being directed horizontally, so the horizontal force is 250*0.6 = 150N. Work = Force * distance = 150N * 6.2m = 930J

4 0

3 years ago