All categories

3 years ago

In regards to mechanical energy, is energy being conserved ?

1 answer:

7 0

If only internal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. ... In these situations, the sum of the kinetic and potential energy is everywhere the same.

You might be interested in

А)

Anastaziya [24]

Answer:

4.5s

Explanation:

u=30m/s

v=50m/s

s=180m

a=constant(given)

By third equation of motion, v

+2as

(50)

=(30)

+2a(180)

m/s

By first equation of motion, v=u+at

50=30+(

=4.5sec

4 0

2 years ago

Humans impact the Earth in good AND bad ways. <br><br>A) True <br><br>B) False

const2013 [10]

Answer:

True

Explanation:

yes we can see that we are helping animals but we create pollution which is very bad

7 0

3 years ago

What is the proper way to start a fire?

Katyanochek1 [597]

Answer:

Explanation:

STEP 1: Gather Your Tools. There's a bit more to building a great campfire than simply placing a few logs in a heap ... If your site has a fire ring, you'll probably have to push the ash and charcoal from ... (Remember, tinder is the really light, quick burning material.) 1. ... Then build a larger teepee of firewood over the kindling.

6 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

3 years ago