All categories

3 years ago

a mass is measured as 6.2kg in one experiment and as 6.20kg in another. in what way do these two measurements differ?

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

They do not differ because 6.2 is the same as 6.20

Explanation:

arlik [135]3 years ago

7 0

They are exactly the same because the numbers of significant figures are the same...adding a zero after a number that is after a decimal does not add another significant figure...for example 6.2 is the same as 6.200000000000

You might be interested in

A ray diagram shows an object placed between 2F and F of a convex lens.The image produced is

nekit [7.7K]

The image produced is magnified and real.

4 0

4 years ago

Read 2 more answers

Please help this is physics!!!

larisa [96]

I’m pretty sure u have it right

8 0

3 years ago

A car of mass 500 kg is moving at a speed of 1.2 m/s. A man pushes the car,

rusak2 [61]

The answer is D I’m not really sure yet

6 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>

7 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago