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3 years ago

The primary purpose of a mirror is to _________ light rays.

1 answer:

8 0

a. reflect (I found this using prior knowledge and process of illumination it can't be absorb cuz you wouldn't see anything it can't be refract because it doesn't reverse the image and it isn't transmit )

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An L-C circuit containing an 82.0-mH inductor and a 1.60-nF capacitor oscillates with a maximum current of 0.800 A . Assuming th

ryzh [129]

Answer:

19.5 mJ

Explanation:

Assuming perfect components without resistance or losses, the circuit should oscillate indefinetly.

The circuit will have a natural pulsation of

$w = \frac{1}{\sqrt{L * C}} = \frac{1}{\sqrt{82e-3 * 1.6e-9}} = 87304 rad/s$

$f = \frac{w}{2\pi} = \frac{87304}{2\pi} = 13895 Hz$

$T = \frac{1}{f} = \frac{1}{13895} = 7.2 \mu s$

So, by the time t = 2.4 ms, 333.33 cycles would have passed

$\frac{2.4 ms}{7.2 \mu s} = \frac{2400 \mu s}{7.2 \mu s} = 333.33$

Therefore it would be at one third after the beginning of the cycle. The circuit would be in an equivalent state as t = (7.2 us)/3 = 2.4us

At t=0 the capacitor is fully charged, so the voltage is maximum and the current is 0. The current will increase towards a maximum of 800 mA at t=T/4, then decreas to 0 at t=T/2, decrease to -800 mA at 3T/4 and go back to 0 at t=T following a sine wave.

The equation of this sine wave would be

$I(t) = I0 * sin(w * t)$

$I(T/3) = 0.8 * sin(w * T/3)$

Since w = 2π/T

w * T = 2π

$I(T/3) = 0.8 * sin(2 \pi /3) = 0.8 * 0.866 = 0.69 A$

The current stored in an inductor is

$E = \frac{1}{2} * L * I^2$

$E = \frac{1}{2} * 82e-3 * 0.69^2 = 0.0195 J = 19.5 mJ$

6 0

3 years ago

An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction

Nesterboy [21]

Answer:

Work = 165670.4 J = 165.67 KJ

Explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

$2as = v_{f}^2 - v_{i}^2\\$

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

$2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2$

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

$F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N$

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

$Work = |F|s\\Work = (4141.76\ N)(40\ m)\\$

8 0

3 years ago

Becomes solid above the melting point

Whitepunk [10]

All liquids become solid, above the melting point!!! Hope this helps!!!

5 0

3 years ago

An ac series circuit contains a resistor of 15 ohms, an inductor of 11.5 mH and a variable capacitor. If the frequency of the ap

SpyIntel [72]

Answer:

C) 20.23 μF

Explanation:

R = resistance of the resistor = 15 ohm

L = inductance of the inductor = 11.5 mH = 0.0115 H

f = resonance frequency achieved = frequency of applied voltage = 330 Hz

C = Capacitance of the capacitor

For resonance to be possible

$2\pi fL = \frac{1}{2\pi fC}$

$2(3.14) (330) (0.0115) = \frac{1}{2 (3.14) (330) C}$

C = 20.23 x 10⁻⁶ F

C = 20.23 μF

5 0

3 years ago

1. In an atom, an electron and a proton are separated by a distance of 1 x 10-8m. a) Find the force on the electron. b) Assuming

aleksandrvk [35]

Answer:

(a) 2.3 x 10^-12 N

(b) 1.6 x 10^5 m/s

Explanation:

q1 = 1.6 x 10^-19 C

q2 = 1.6 x 10^-19 C

r = 1 x 10^-8 m

(a) the electrostatic force is given by

F = K q1 x q2 / r^2

F = ( 9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (1 x 10-^-8)^2

F = 2.3 x 10^-12 N

The force is attractive in nature because the nature of charge on electron and proton s opposite to each other.

(b) The electrostatic force is balanced by the centripetal force actig on the electron.

F = mv^2 / r

where, r be the radius of orbit and v be the velocity of electron in the orbit.

2.3 x 10^-12 = (9.1 x 10^-31 x v^2) / (1 x 10^-8)

v = 1.6 x 10^5 m/s

6 0

3 years ago