When an electron absorbs energy it jumps to what state?

If the western component of the wind is half of the south, what is the angle of the wind with the south?

den301095 [7]

Answer:

This above a triangle that models our situation.

Explanation:

We have a two componens., since we have a western componet and southern component. One travel in a southern direction. and the other travel in the west.

Let the component that travel in the south be the length of a.

According to the problem, the westard component is half of that so let that length be a/2.

Now we must find the angle of the wind in the South.

This means that what is angle that is opposite of the western componet because that angle is the most southward angle. So know we apply the tan property.

$\tan(x) = \frac{opp}{adj}$

Our side opposite of the angle we trying to find is the western component and the side adjacent to it is the southern component. Also remeber since western and Southern negative displacements, we have

$\tan(x) = \frac{ - \frac{a}{2} }{ - a}$

$\tan(x) = - \frac{a}{2} \times - \frac{1}{a} = \frac{1}{2}$

Now we take the arctan or inverse tan of 1/2.

$\tan {}^{ - 1} ( \frac{1}{2} ) = 26.57$

3 0

2 years ago

Read 2 more answers

A device used to increase or decrease the emf in the second of two unconnected coils is a

grigory [225]

The answer is transformer. They utilize electromagnetic induction to generate current. This is only possible in alternating current due to the differential increase and decrease of electrical current that induces changes in magnetic flux in the coil. This varies the magnetic flux of the primary coil that generates current in the secondary coil.

4 0

3 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago

the public is not yet able to purchase hydrogen fuel cell powered cars because engineers have to determine

grigory [225]

Answer: This would more a practical feedback than scientific.

Explanation: Despite of the protocols and processes scientists need to develop in order to stablish a new source of energy or daily life option, logistic issues are one of the most important things on the "to do list".

Would it be very expensive to acquire this kind of fuel? Is it efficient? Is it dangerous to human life or the environment? What about the negative reactions this brand new fuel could have in society?.

3 0

4 years ago

g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a

otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$