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2 years ago

The decibel scale is used to measure the perceived loudness of a sound. a. True b. False

Physics

1 answer:

Mademuasel [1]2 years ago

8 0

True because when you look at the scale you see that some louder things are higher on it than others

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Dimensional Analysis : 3 days to seconds

LuckyWell [14K]

Hi,

To convert 3 days to seconds write this.

1h = 3600s
24h = 3600 · 24 = 86400
3 days = 3 · 86400 = 259200sec

Hope this helps.
r3t40

8 0

2 years ago

Read 2 more answers

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

A small cart travels 20 meters to the right in 10 seconds calculate the velocity

stira [4]

The velocity would be 10

3 0

2 years ago

Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1

max2010maxim [7]

Answer:

2.5 m/s

Explanation:

Mechanical energy is the sum of the potential and kinetic energy.

E = PE + KE

E = mgh + ½mv²

172.1 J = (7.26 kg) (9.8 m/s²) (2.1 m) + ½ (7.26 kg) v²

v = 2.5 m/s

7 0

2 years ago

Read 2 more answers