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3 years ago

Weight and mass of an object are the same by

Engineering

1 answer:

dangina [55]3 years ago

6 0

Answer:

Explanation:

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. ... Weight usually is denoted by W. Weight is mass multiplied by the acceleration of gravity (g).

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(1) 1. (15 points/ 3 points each) (a) Draw the binary search tree that is created if the following numbers are inserted in the t

Alex777 [14]

Answer:

The binary search tree BST that is created is shown in the figure in the attached file

The missing part of the question is to draw the balanced binary search tree containing the same numbers given in the question.

6 0

3 years ago

A murder in a downtown office building has been widely publicized. You’re a police detective and receive a phone call from a dig

BaLLatris [955]

Answer:

Considering the plain view doctrine, which is an exception to the warrant requirement of the Fourth Amendment, is applied by law enforcement officers and courts who can seize evidence of a crime without a warrant, if the officer observes the evidence in plain view.

Explanation:

For any digital information related to a murder case that has been seized under the plain view doctrine to be used to convict you of a crime, has to comply with three conditions:

1. The digital evidence must be in out in the open, and easily observable by the officer, this is what "plain view" refers to.

2. The officer must have a legal right to be where he got the information related to the case.

3. The 'incriminating' character of the information must be a clear hint of the murder to fall under the plain view doctrine and the officer´s experience will help him determining whether the information is evidence or not, upon probable cause related to a crime.

7 0

3 years ago

Read 2 more answers

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

3 years ago

One good way to improve your gas Milage is to ___.

VashaNatasha [74]

Answer: B

Explanation:

One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.

Hope this helps!

5 0

3 years ago

Read 2 more answers

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following

torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches $5\times 10^5$

$(a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\$

$(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e$ $(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}$

7 0

3 years ago