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3 years ago

Need help please help

Physics

1 answer:

deff fn [24]3 years ago

7 0

∠BAC= 180-110 (angles on a straight line=180)

∠BAC=70

Triangle ABC is isoceles

∠ABC=∠ACB

Sum of angles in a triangle = 180

180-70=110

110/2=55

x=55

∠BCA= 55 (bottom angles are equal)

180-55=125

y=125

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A 30.0-Î¼F capacitor is connected to a 49.0-Î© resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms

Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

$Z=\sqrt{R^2+X_{c}^2}$ .....(I)

We need to calculate the value of $X_{c}$

Using formula of $X_{c}$

$X_{c}=\dfrac{1}{2\pi f c}$

$X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}$

$X_{c}=88.42\ \Omega$

Put the value of $X_{c}$ into the formula of impedance

$Z=\sqrt{(49.0)^2+(88.42)^2}$

$Z=101.08\ \Omega$

(a). We need to calculate the rms current in the circuit

Using formula of rms current

$I_{rms}=\dfrac{V}{Z}$

$I_{rms}=\dfrac{30.0}{101.08}$

$I_{rms}=0.30\ A$

The rms current in the circuit is 0.30 A.

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

Using formula of rms voltage

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

Put the value into the formula

$V_{rms}=0.30\times49.0$

$V_{rms}=14.7\ V$

The rms voltage drop across the resistor is 14.7 V

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

Using formula of rms voltage

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

$V_{rms}=0.30\times88.42$

$V_{rms}=26.53\ V$

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

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3 years ago

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Answer:

b. K = U

Explanation:

In this case you have that there are no non-conservative forces over the pendulum. Hence, the total mechanical energy if the pendulum must conserve. You take into account that the potential and kinetic energy of a pendulum are given by:

$E_T=K+U=\frac{1}{2}mv^2+mgh$

m: mass of the pendulum

h: height of the pendulum

v: speed

In each moment of the trajectory of the pendulum ET does not change.

You have that, at the lowest point U=0J and for the highest point K=0J. For that point K=ET and U=ET respectively.

Hence, for a mid-way between those points it is necessary that:

b. K = U

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The astrometric technique of planet detection works best for

denpristay [2]

The astrometric technique of planet detection works best for massive planets around nearby stars.

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

nexus9112 [7]

(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is

a = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²

so that the average net force on the man during the landing is

F = (70.8 kg) a ≈ 225,000 N

i.e. with magnitude 225,000 N.

(b) With knees bent, the man has an average acceleration of

a = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²

and hence an average net force of

F = (70.8 kg) a ≈ 3050 N

(c) The net force on the man is

∑ F = n - w = m a

where

n = magnitude of the normal force, i.e. the force of the ground pushing up on the man

w = the man's weight, m g ≈ 694 N

m = the man's mass, 70.8 kg

g = mag. of the acceleration due to gravity, 9.80 m/s²

a = the man's acceleration

Using the acceleration in part (b), we have

n = m g + m a = m (g + a)

n = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N

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does a Neon atom emits specific frequencies of light that correspond to decreases in energy of its electrons.

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Answer:

Yes, the frequency of light emitted is a property of the difference between the levels of energy of its electrons.

Explanation:

Neon atom is a noble gas which glows when its electrons de-excites after absorbing energy.

Niels Bohr postulated that the energy level in all atoms are quantized, thus electrons do not exist in-between two levels. When electrons in the Neon atom are excited, this increase in energy causes them to jump to a higher energy levels. On de-excitation, the electrons drops to their initial level releasing the absorbed energy in the form of a photon.

The photon emitted has a frequency that is directly proportional to the energy change in the electron.

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