If s is the specific heat capacity of 4kg water, what is the specific heat capacity of 16kg

Which of the following statements best describes the balls motion ?

Schach [20]

Answer:

C

Explanation:

8 0

3 years ago

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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

So

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago

Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.

serg [7]

Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..

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3 years ago

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About 2% of our solar nebula consisted of elements besides hydrogen and helium. However, the very first generation of star syste

Mariulka [41]

Answer:

Answer

Explanation:

The very first generation of star systems in the universe probably consisted only of hydrogen and helium. This means that,

There were no comets or asteroids in these first-generation star systems. As the comets and asteroids were the only soruce of elements other than Hydrogen and Heliums ( As starts mainly consists of helium and hydrogen).

3 0

3 years ago