What is the KE of a 27 kg mass moving at 3m/s?

Which factor affects elastic potential energy but not gravitational potential energy?

Vika [28.1K]

Answer:

The correct answer is the spring constant (k)

Explanation:

Elastic potential energy: Epe = ½ * k * Δx²

k = spring constant

Δx = displacement

Gravitational ptential energy: Ep = m*g*h

g = (acceleration due to gravity)

m = mass

h = relative height of an object to some reference point

3 0

3 years ago

Read 2 more answers

Your digestive system has three main functions digestion, absorption and elimination. in which organ does absorption begin?

xeze [42]

The Answer Is The Small Intestine

3 0

3 years ago

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Many people believe that a vacuum created inside a

pashok25 [27]

Answer:Due to the pressure difference created by rotating fans.

Explanation:

In most of the vacuum cleaners, there is an area which is of disc shape and it is in right next to the motor. There are several fans within the disc that spin at a very high velocity.

The blades will push the air outside of the disk.There is no air in inside of the disc and air pressure creates which pushes air inside the disk to replace the missing air.

So motor is pushing the air outside and to maintain this pressure the air is pushing toward inside with dirt.

3 0

3 years ago

the book has a mass of 2.5 kg. What net force must act on the book to mak it accelerate to the left at a rate of 7.0m/s2?

Reil [10]

Answer:

17.5 N

Explanation:

<h2>Given :</h2>

Mass (m) = 2.5 kg
Acceleration (a) = 7.0 m/s²

<h2>To calculate :</h2>

Force exerted (F)

<h2>Calculation :</h2>

→ F = (2.5 × 7.0) N

→ F = 25/10 × 7 N

→ F = 5/2 × 7 N

→ F = (5 × 7)/2 N

→ F = 35/2 N

→ <u>F</u><u> </u><u>=</u><u> </u><u>1</u><u>7</u><u>.</u><u>5</u><u> </u><u>N</u><u> </u><u>towards</u><u> </u><u>left</u>

Hence, 17.5 N of net force must act on the book to make it accelerate to the left.

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

4 years ago