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Neko [114]
3 years ago
11

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Physics
1 answer:
noname [10]3 years ago
8 0

Answer:

i dont think anything will happen so maybe A: the wall is larger than you.

Explanation:

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You are asked to find a set of dimensionless parameters to organize data from a laboratory experiment, in which a tank is draine
mihalych1998 [28]

Answer:

Step 1

Given: A tank is drained through an orifice from initial liquid level (h₀). The time(r), to  drain the tank, depends on tank diameter (D), orifice diameter (d), acceleration due to gravity (g), liquid density (ρ) and liquid viscosity  (μ).

r = f (h₀, D, d, g, ρ, μ)

Step 2

There are totally seven parameters r, h₀, D, d, g, ρ and μ

∴ n = 7

The primary variables are M, L, T

∴ m = 3

So, the number of repeating variables are, r = 3

Select the three repeating variables are ρ, d, g

Then according to Buckingham, there will be n-m (⇒ 7-3 = 4) dimensionless groups given by π₁, π₂, π₃ and π₄

So, The number of π terms = 4

Explanation:

See attached images for step 3 to 7

6 0
3 years ago
true or false. because the speed of an object can change from one instant to the next, dividing the distance covered by the time
scoundrel [369]

true. Because the speed of an object can change from one instant to the next, dividing the total distance covered by the time of travel gives. average speed.

3 0
3 years ago
Read 2 more answers
During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t^2 to s
Mila [183]

Answer:

70560 m

Explanation:

The formula to calculate the distance travelled during a free fall motion is

d=-\frac{1}{2}gt^2

where

d is the distance travelled

g = -9.8 m/s^2 is the acceleration due to gravity

t is the time

In this situation,

t = 120 s

Therefore the distance travelled after 120 s is

d=-\frac{1}{2}(-9.8)(120)^2=70560 m

6 0
4 years ago
The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic
Drupady [299]

a) Density at 100 degrees: 1.34\cdot 10^4 kg/m^3

Explanation:

The density of mercury at 0 degrees is d=1.36\cdot 10^4 kg/m^3

Let's take 1 kg of mercury. Its volume at 0 degrees is

V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3

The formula to calculate the volumetric expansion of the mercury is:

\Delta V= \alpha V \Delta T

where

\alpha=180\cdot 10^{-6} K^{-1} is the cubic expansivity of mercury

V is the initial volume

\Delta T is the increase in temperature

In this part of the problem, \Delta T=100 C-0 C=100 C=100 K

So, the expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3


b) Density at 22 degrees: 1.355\cdot 10^4 kg/m^3

We can apply the same formula we used before, the only difference here is that the increase in temperature is

\Delta T=22 C-0 C=22 C=22 K

And the volumetric expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3


8 0
3 years ago
Help me with Economics please and thank you the question is going to be down in a attached file while the answers on here
Tresset [83]
I think is A or B it depends on like what the trying to answer
6 0
3 years ago
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