How do you label Time= energy • power

1 answer:

5 0

First of all, that equation is not correct, which may be the reason
that you're having trouble assigning units to the quantities.

Power is defined as [energy / time], so [Energy] = [ power x time ],
and
[Time] = [ energy / power ].

Unit-wise, these equations are correct just as they appear here,
with no proportionality constants or conversion factors, when ...

[ Power ] = watts
[ Energy ] = joules
[ Time ] = seconds .

You might be interested in

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.8 t ?

defon

The magnetic force acting on the proton is
$F=qvB \sin \theta$
where
q is the proton charge
v is its speed
B is the intensity of the magnetic field
$\theta$ is the angle between the direction of v and B; since the proton is moving perpendicular to the magnetic field, $\theta=90^{\circ}$ and $\sin \theta=1$ , so the force becomes
$F=qvB$

this force provides the centripetal force that keeps the proton in circular motion:
$m \frac{v^2}{r} = q v B$
where the term on the left is the centripetal force, with
m being the mass of the proton
r the radius of its orbit

Re-arranging the previous equation, we can find the radius of the proton's orbit:
$r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m$

And now we can calculate the centripetal acceleration of the proton, which is given by
$a_c = \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2$

7 0

3 years ago

A roulette wheel with a 1.0m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions ( 220 rad ) after

Alex17521 [72]

Max ang. speed(u) = 18 rad/s
final ang. speed(v) = 0
ang. displacement(s) = 220 rad

ang. acceleration = (v^2 - u^2)/2s = -18^2 / 2*220 = -0.7364 rad/s^2

v = u +at
0 = 18 - 0.7364t
t = 18/0.7364
t = 24.44 seconds

4 0

2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$