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astraxan [27]
3 years ago
13

While doing an experiment you measure the length of an object to be 32.5cm. The measuring device that you are using measures to

a millimeter. How can you decrease the error associated with your measurements
Physics
1 answer:
Allushta [10]3 years ago
3 0

Answer and Explanation:

By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise  and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.

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A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav
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Answer:

The tension in the string is T = 1.49*10^{-6}N.

Explanation:

For a string with tension T and linear density \mu_d carrying a transverse wave at speed v it is true that

v = \sqrt{\dfrac{T}{\mu_d} }

solving for T we get:

T = \dfrac{v^2}{\mu_d}.

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s

And it's linear density (mass per unit length) is

\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m

Therefore, the tension in the cord is

T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.

\boxed{T = 1.5*10^{-6}N}

or in micro newtons

T =1.5\mu N

4 0
3 years ago
The derived unit for density are g/cm3.<br><br> True<br> False
anzhelika [568]
True, the measurement shown is a derived unit.
6 0
3 years ago
How Can the human ear detect the sound
liq [111]

Explanation:

The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear. These bones are called the malleus, incus, and stapes.

8 0
2 years ago
Science fair project
victus00 [196]

For Science fair project you can do convection current

Explanation:

Show how to do the process

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8 0
3 years ago
Read 2 more answers
A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

4 0
3 years ago
Read 2 more answers
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