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3 years ago

At what temperature did the substance in the test tube into a solid?

Physics

1 answer:

katrin2010 [14]3 years ago

6 0

What are the questions choices?

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Help ASAP thanks !!!

Ann [662]

Answer:

true

Explanation:

3 0

3 years ago

a star has a distance of 30 parsecs and a apparent magnitude of 3 --what would its absolute magnitude be?

Alborosie

The absolute magnitude of the star would be +5.

8 0

1 year ago

Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0

3 years ago

It is important for physicians to be _____ patient use of alternative therapies.

V125BC [204]

It is important for physicians to be Respectful of patient use of alternative therapies

Alternative therapies usually have not gain a scientific approval from the scientific community, but there are some therapies that show positive results even though it is still has not proved by studies or researches. So it's important for physicians to be respectful if the patient choose to do it

3 0

3 years ago

A circuit contains a 305 ohm resistor, a 1.1 micro Farad capacitor, and a 42 mH inductor. At resonance, the impedance is determi

r-ruslan [8.4K]

Answer:

The resistance of the inductor at resonance is 258.76 ohms.

Explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

$\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L = \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }$

Where;

F₀ is the resonance frequency

$F_0 = \frac{1}{2\pi\sqrt{LC} } \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^{-6})} }\\\\F_0 =980.4 \ Hz$

The inductive reactance is given by;

$X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms$

Therefore, the resistance of the inductor at resonance is 258.76 ohms.

5 0

3 years ago