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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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A 150 kg car hits a wall with 45 N of force, what was its acceleration?

alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: $a = \frac{F}{m}$

Solution: To find a, divide the force by the weight

A = F ÷ m

= 45 ÷ 150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

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3 years ago

A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when

Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

$v^2-u^2=2ad$

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

$a=-8.0 m/s^2$ is the acceleration

Solving for u, we find the initial velocity:

$u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s$

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3 years ago

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

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3 years ago

A 5000-kg freight car runs into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. What

Vitek1552 [10]

Answer:

C) 6 m/s

Explanation:

Given that

m₁=5000 kg

The initial velocity of 5000 kg car =u₁

m₂=10,000 kg

The initial velocity of 10000 kg car =u₂ = 0 m/s

After collision the final speed of the both car,v = 2 m/s

There is no any external force on the system that is why linear momentum will be conserved.

Linear momentum P = m v

m₁u₁ + m₂u₂ = (m₂ + m₁) v

5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2

5000 x u₁ = 15000 x 2

5 x u₁ = 15 x 2

u₁ = 6 m/s

Therefore the answer is C.

C) 6 m/s

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3 years ago

Which of the following statements best describes the relationship between force and work?

viktelen [127]

a)., b)., and c). are completely false.
There isn't a grain of truth among them.

In Physics, the technical definition of 'Work' is (force) times (distance).

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3 years ago