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3 years ago

Your body is receiving backround radiation all the tim. Explain

1 answer:

3 0

The uniform microwave radiation remaining from the Big Bang.

So, your body is always having background radiation and that means space!

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an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo

Liono4ka [1.6K]

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W = 87.08 [6y⁴/4 - y³/3]

W = 87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

To know more about work done:

brainly.com/question/16650139

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7 0

2 years ago

Particle motion in surface waves is __________ motion.

Readme [11.4K]

Answer:

A combination of longitudinal & transverse

Explanation:

6 0

3 years ago

Read 2 more answers

59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determin

Daniel [21]

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

$x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)$

$y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)$

From Eqn(2), we see that

$N = mg\cos 25°\;\;\;\;\;\;\;(3)$

so using Eqn(3) on Eqn(1), we get

$mg\sin 25° - \mu_kmg\cos 25° = ma$

Solving for the acceleration, we see that

$a = g(\sin 25° - \mu_k\cos 25°)$

$\;\;\;\;= 2.45\:\text{m/s}^2$

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

$v^2 = v_0^2 + 2ax$

Since the crate started from rest, $v_0 = 0.$ Thus our equation reduces to

$v^2 = 2ax \Rightarrow v = \sqrt{2ax}$

$v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}$

$\;\;\;\;= 6.32\:\text{m/s}$

6 0

3 years ago

Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I

mezya [45]

The problem is basically asking us to find a way to find the sound intensity I, in terms dependent on the sound level and the reference intensity $I_0$ .For this purpose we can start from the unit used in the scale logarithmic decibel, that is

$\beta = 10log_{10}\frac{I}{I_0}$

Where

I = Acoustic intensity on the linear scale

$I_0 =$ Hearing threshold

Using the logarithmic properties of the exponents the above expression can be described as:

$(\frac{I}{I_0})^{10} = 10^{\beta}$

$I = I_0 10^{\frac{\beta}{10}} \righarrow$ that is the expression or technique to find the intensity of sound.

8 0

3 years ago

The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is suf

kupik [55]

Answer:

(a) $I_{1}=3.2*10^{-3}W/m^{2}$

(b) $\beta =95dB$

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

$\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}$

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

$\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB$

8 0

3 years ago