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3 years ago

Which has greater accuracy, the 10 mL graduated cylinder or the 50 mL graduated cylinder

2 answers:

6 0

You can estimate one more digit past the smallest division on the measuring device. If you look at a 10mL graduated cylinder, for example, the smallest graduation is tenth of a milliliter (0.1mL). That means when you read the volume, you can estimate to the hundredths place (0.01mL).

vampirchik [111]3 years ago

5 0

The 10ml would be more accurate.

You might be interested in

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

Vesnalui [34]

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere= $\rho$

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

$Density=\frac{mass}{volume}$

$Mass=Density\times volume=Constant$

Therefore, $\rho_1 V_1=\rho_2V_2$

Volume of sphere= $\frac{4}{3}\pi r^3$

Using the formula

$\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3$

$\rho R^3=\rho_2\times \frac{R^3}{8}$

$\rho_2=8\rho$

Hence, the density of the compressed sphere= $8\rho$

Option a is correct.

7 0

2 years ago

Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache

HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f is when n = 1. So,

f = 1/2L√(T/μ) = 1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f' is when n = 3. So,

f' = 3/2L√(T/μ) = 3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0

2 years ago

the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce

anastassius [24]

Answer:

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So $24=32+a\times 4$

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

6 0

2 years ago

NEED ANSWER ASAP!!!!!!

olga55 [171]

Answer:

yes. why do you need this answered asap? lol

5 0

3 years ago

Read 2 more answers

Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

xxMikexx [17]

Answer:

Frequency of the light will be equal to $6.97\times 10^{1}Hz$

Explanation:

We have given wavelength of the light $\lambda =430nm=430\times 10^{-9}m$

Velocity of light is equal to $v=3\times 10^8m/sec$

We have to find the frequency of light

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency of light

So frequency of light will be equal to $f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz$

So frequency of the light will be equal to $6.97\times 10^{1}Hz$

5 0

3 years ago