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2 years ago

Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above

Physics

1 answer:

Elanso [62]2 years ago

6 0

Answer:

I think is 2

Explanation:

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Where is the centre of mass of a system of two particles is situated?

Sever21 [200]

Answer:

In a two particle system, the center of mass lies on the center of the line joining the two particles.

4 0

2 years ago

QC In ideal flow, a liquid of density 850 kg / m³ moves from a horizontal tube of radius 1.00cm into a second horizontal tube of

kvv77 [185]

The volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

According to bernaulli's theorem theorem

P+1/2pV²+pgy = constant

where p fluid density

g is acceleration due to gravity, pressure at elevation,v is Velocity at elevation ,y is height of elevation.

As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

4 0

1 year ago

The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

2 years ago

Please I need HELP WITH THIS QUESTION!!!

JulsSmile [24]

Oil, grease and dry lubricants

8 0

2 years ago

An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd

Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

= 1.07 × 10³(1 - [0.024]²)

= 1.07 × 10³(1 - 0.000576)

= 1.07 × 10³(0.999424)

= 1069.38 gallons

7 0

3 years ago