All categories

4 years ago

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

Physics

1 answer:

wolverine [178]4 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.

You might be interested in

Organic macromolecules called _______ are insoluble in water

NikAS [45]

Answer:

lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.

3 0

3 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

3 years ago

Read 2 more answers

In chemical reactions, energy is absorbed or released in the form of:

disa [49]

Answer:

Heat

Explanation:

Because chemical energy is stored, it is a form of potential energy. When a chemical reaction takes place, the stored chemical energy is released. Heat is often produced as a by-product of a chemical reaction – this is called an exothermic reaction.

Hope this helped.

3 0

3 years ago

Read 2 more answers

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

We start making summation of Moments in A,

$\sum M_A = 0$