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3 years ago

11

Evelynn is measuring the pitch of a piano note. What unit of measurement is she most likely recording her value in? hertz decibe

ls frequency amplitude

2 answers:

Gnom [1K]3 years ago

5 0

The unit of measurement for pitch is Hertz, Hz.

german3 years ago

4 0

Answer:

above me is right

Explanation:

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Please help ASAP for the fill the gap sentences. The brackets are for the energy stores..... thank you in advance.

lesya692 [45]

The answer is c or h

4 0

3 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago

The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty

nydimaria [60]

Answer:

Uncertainty in position of the bullet is $\Delta x=1.07\times 10^{-33}\ m$

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

$p=mv$

$p=0.035\times 709=24.81\ kg-m/s$

Uncertainty in momentum is,

$\Delta p=0.2\%\ of\ 24.81$

$\Delta p=0.049$

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x\geq \dfrac{h}{4\pi}$

$\Delta x=\dfrac{h}{4\pi \Delta p}$

$\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}$

$\Delta x=1.07\times 10^{-33}\ m$

Hence, this is the required solution.

7 0

3 years ago

The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

sattari [20]

Answer:

k = 11,564 N / m, w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

let's apply the equilibrium condition at this point

Axis y

W_{y} - Fr = 0

Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

sin 46 = $W_{y}$ / W

W_{y} = W sin 46

we substitute

mg sin 46 = k y

k = mg / y sin 46

If the length of the bar is L

sin 46 = y / L

y = L sin46

we substitute

k = mg / L sin 46 sin 46

k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

k = 1.18 9.8 / 1

k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

Em₀ = U = mgy

end point. Point at 30º

$Em_{f}$ = K -Ke = ½ I w² - ½ k y²

em₀ = Em_{f}

mgy = ½ I w² - ½ k y²

w = √ (mgy + ½ ky²) 2 / I

the height by 30º

sin 30 = y / L

y = L sin 30

y = 0.5 m

the moment of inertia of a bar that rotates at one end is

I = ⅓ mL 2

I = ½ 1.18 12

I = 0.3933 kg m²

let's calculate

w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

w = 6.06 rad / s

7 0

3 years ago

A charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a th

Mila [183]

Answer:

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

Explanation:

Position of charge 3q is x = 0

position of charge -2q is x = a

so here we know that

when two charges are of opposite nature then the electric field will be zero on the line joining the two charges at the position near to smaller magnitude charge

So here the electric field will be zero if the field due to 3q is counterbalanced by field due to -2q

so here we can say

$\frac{k(3q)}{(r+a)^2} + \frac{k(-2q)}{r^2} = 0$

$\frac{3}{(r + a)^2} = \frac{2}{r^2}$

$\frac{r+a}{r} = \sqrt{\frac{3}{2}}$

$1 + \frac{a}{r} =\sqrt{\frac{3}{2}}$

$\frac{a}{r} = \frac{\sqrt3 - \sqrt2}{\sqrt2}$

so we will have

$r = \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

so the x coordinate of this position is given as

$x = a + \frac{a\sqrt2}{\sqrt3 - \sqrt2}$

6 0

3 years ago