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3 years ago

10. How do you use a triple-beam balance? Fill in the blanks.

Physics

1 answer:

wolverine [178]3 years ago

5 0

Answer:

1st – Place the film canister on the scale.

2nd – Slide the large weight to the right until the arm drops below the line and then move it back one notch.

3rd – Repeat this process with the top weight. When the arm moves below the line, back it up one groove.

4th – Slide the small weight on the front beam until the lines match up.

5th – Add the amounts on each beam to find the total mass to the nearest tenth of a gram.

Explanation:

The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.

The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.

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In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.

galben [10]

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

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2 years ago

If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if

umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

$P'=P+\rho gh$

Put the value into the formula

$P'=1.01\times10^{5}+1025\times9.8\times10$

$P'=201450\ Pa$

We need to calculate the volume at the surface

Using equation of ideal gas

$PV= RT$

So, for both condition

$PV=P'V'$

Put the value into the formula

$V=\dfrac{201450\times5.5}{1.01\times10^{5}}$

$V=10.97\ L$

Hence, The volume at the surface is 10.97 L.

3 0

2 years ago

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Cho hai điện tích q1=q2=8.10^-7 C đặt cách nhau 5cm. Xác định cường độ điện trường tại điểm:

Aleksandr [31]

Answer: b

Explanation:

5 0

2 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

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3 years ago

Where should you place the values of the independent variable when constructing a data table?

kobusy [5.1K]

On the y-axis (the bottom of the table) hope this helps

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3 years ago

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