All categories

3 years ago

Công thức dùng tính tần số dao động của con lắc lò xò

Physics

1 answer:

tino4ka555 [31]3 years ago

6 0

Answer:

f = 1 / [ 2*pi*R*C*Sqrt(2*N)

Explanation:

f = 1 / [ 2*pi*R*C*Sqrt(2*N) is the formula. Sorry I answered in English.

You might be interested in

Most offshore drilling occurs:

MAXImum [283]

In the Atlantic Ocean

5 0

3 years ago

A parallel-plate capacitor has an area of 4.59 cm2, and the plates are separated by 1.28 mm with air between them. it stores a c

nalin [4]

<span>a)
Capacitance = k x ε° x area / separation
ε° = 8.854 10^-12 F/ m
k = 2.4max
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86
area = 145 cm2 = 0.0145 m2
separation = 1.27 cm 0.0127 m

C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>

7 0

3 years ago

The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot

KiRa [710]

Answer:

$T_{f}$ = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

Q₁ = 2.00 10⁻³ 2.26 10⁶

Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

Q coffee = M $c_{e}$ ( $T_{f}$ - $T_{i}$ )

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

Qc = - Q₁

M ce ( $T_{f}$ - $T_{i}$ ) = - Q₁

The coffee dough left in the cup after evaporation is

M = 250 -2 = 248 g = 0.248 kg

$T_{f}$ -Ti = -Q1 / M $c_{e}$

$T_{f}$ = Ti - Q1 / M $c_{e}$

Since coffee is essentially water, let's use the specific heat of water,

$c_{e}$ = 4186 J / kg ºC

Let's calculate

$T_{f}$ = 90.0 - 4.52 103 / (0.248 4.186 103)

$T_{f}$ = 90- 4.35

$T_{f}$ = 85.65 ° C

$T_{f}$ = 85.7 ° C

5 0

3 years ago

A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field

arsen [322]

magnetic field due to a finite straight conductor is given by

$B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)$

here since it forms an equilateral triangle so we will have

$\theta_1 = \theta_2 = 60 degre$

also the perpendicular distance of the point from the wire is

$r = \frac{a}{2\sqrt3}$

now from the above equation magnetic field due to one wire is given by

$B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)$

$B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)$

$B = \frac{3\mu_0 i}{2\pi a}$

now since in equilateral triangle there are three such wires so net magnetic field will be

$B = \frac{9\mu_0 i}{2\pi a}$

5 0

3 years ago

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

$P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,$

where $h_1$ and $h_2$ are the height of the column of oil and the unkown liquid, respectively. Writing for $\gamma_{unk}$ , we have

$\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.$

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0

3 years ago