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3 years ago

Which metal has the lowest atomic weight

Physics

2 answers:

Ray Of Light [21]3 years ago

7 0

Lithium, Li = 6.94 :)

Ber [7]3 years ago

7 0

Lithium, it is around 6.4 in atomic weight.

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Images produced by a cathode ray tube result from induced emf True or false

3241004551 [841]

I'm not completely sure but i think it's false

8 0

3 years ago

In a certain region of space, the electric field is constant in direction (say horizontal, in the x direction), but its magnitud

horsena [70]

Answer:

8.3 x 10⁻⁷ C

Explanation:

Electric flux will enter the face at x=0 and exit at face x= 25 m

On the other faces , field lines are parallel so no flux will enter or exit .

Flux entering the face at x = 0

= electric field x face area

= 560 x 25 x 25 = 350000 weber

Flux exiting the face at x = 25

= 410 x 25 x25

= 256250 weber

Net flux exiting from cube ( closed face )

350000 - 256250 = 93750 web

Apply gauss'es theorem

Q / ε = Flux coming out

Q is charge inside the closed cube

Q / ε = 93750

Q = 8.85 x 10⁻¹² x 93750

= 8.3 x 10⁻⁷ C

7 0

3 years ago

For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz.

Natasha_Volkova [10]

Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance $C = 25.0\ \mu C$

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

$\omega = 2\pi f$

Put the value into the formula

$\omega =2\times3.14\times60.0$

$\omega=376.8\ rad/s$

(a). We need to calculate the value of $X_{L}$

Using formula of $X_{L}$

$X_{L}=\omega\times L$

Put the value into the formula

$X_{L}=376.8\times55.0\times10^{-3}$

$X_{L}=20.724\ \Omega$

(b). We need to calculate the value of $X_{L}$

Using formula of $X_{C}$

$X_{C}=\dfrac{1}{\omega C}$

$X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}$

$X_{C}=106.16\ \Omega$

(c). We need to calculate the value of Z

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

Put the value into the formula

$Z=\sqrt{75.0^2+(20.724-106.16)^2}$

$Z=113.68\ \Omega$

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{12.0}{75.0}$

$I=0.16\ A$

Using formula of rms current

$I_{rms}=\dfrac{I_{0}}{\sqrt{2}}$

$I_{rms}=\dfrac{0.16}{\sqrt{2}}$

$I_{rms}=0.113\ A$

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

$V_{rms}=I_{rms}\times R$

$V_{rms}=0.113\times75.0$

$V_{rms}=8.475\ V$

(f). We need to calculate the rms voltage across the inductor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{L}$

$V_{rms}=0.113\times20.724$

$V_{rms}=2.342\ V$

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

$V_{rms}=I_{rms}\times X_{C}$

$V_{rms}=0.113\times106.16$

$V_{rms}=11.99\ V$

(h). We need to calculate the dissipated power by the circuit

Using formula of dissipated power

$P=RI^2$

Put the value into the formula

$P=75.0\times0.113^2$

$P=0.958\ W$

Hence, This is the required solution.

3 0

3 years ago

What is the potential energy of a spring that is stretched 0.15 m from equilibrium and has a spring constant of 0.55 N/m?

yan [13]

Explanation:

EE = ½ kx²

EE = ½ (0.55 N/m) (0.15 m)²

EE = 0.62 J

8 0

3 years ago

How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?

AleksandrR [38]

Answer:

270Joues

Explanation:

Step one:

given data

Force F= 45N

distance moved d= 6m

Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

WD= 45*6

WD=270Joues

7 0

3 years ago