Which three elements have strong magnetic properties?

An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb

ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

= 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0

3 years ago

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

Luba_88 [7]

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

$E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}$

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL² will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

3 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f

Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 ) ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66² - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

4 0

2 years ago

Could someone answer this?

Nadusha1986 [10]

Answer:

Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

Now,

The maximum power of each resistance is 16 W

The 4Ω resistor is linked in series with the circuit.

so, P o w e r = I

two

R, here i is the current through the resistor resistor R

1 6 = I

two

∗ 4 Ω

i = 2A

Now 2A passes through parallel resistors of 8Ω resistance.

we know that, in parallel, the potential difference must be constant,

the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is

2 A

two

.

= 1 A

So finally the current through the 4Ω resistor = 2 A

current through each 8Ω resistor = 1 A

Explanation:

I hope this answer has helped you

6 0

3 years ago