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3 years ago

Calculate the density of water in kg/m3. Express your answer in kilograms per cubic meter using three significant figures.

Physics

1 answer:

Svet_ta [14]3 years ago

8 0

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, the density of water is generally known to be 997 kg/m³. The formula used to determine density is mass ÷ volume. In the case of the density of water, the mass is measured in kilograms (kg) while the volume is measured in cubic meter (m³).

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Newton's 2nd law describes the relation between

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It describes the relationship between force and acceleration

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3 years ago

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

3 years ago

What is meant by specific latent heat of fussion

Lunna [17]

Answer:

Boiling point of water is 100 deg C. It takes about 540 cal to change 1 g of water at the boiling point to 1 g of steam at 100 deg C.

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2 years ago

A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0

3 years ago