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3 years ago

5

A database that is flexible and easy to use was recently installed. the data is stored in rows and columns much like a spreadshe

et. what type of database must have been installed?

1 answer:

andre [41]3 years ago

7 0

The correct answer is Relational Database

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A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0Â° above the horizontal. it strike

Anni [7]

<span>x = 129.9 m y = 30.9 m First, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m</span>

8 0

3 years ago

A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i

Lubov Fominskaja [6]

Answer:

11.8 m/s

Explanation:

At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).

Sum of forces in the centripetal direction:

∑F = ma

mg − N = m v²/r

At the maximum speed, the normal force is 0.

mg = m v²/r

g = v²/r

v = √(gr)

v = √(9.8 m/s² × 14.2 m)

v = 11.8 m/s

3 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

3 years ago

Atoms of which pair of elements will form ionic bonds in a compound?

pishuonlain [190]

Answer:

Be and Cl

Explanation:

This pair will form an ionic bond because Be is a metal while Cl is a nonmetal.'

Ionic bonds are bonds between nonmetals and metals.

5 0

2 years ago

"Caleb misjudges while parking, and bumps his car into a brick wall at 3 MPH. His bumper brings the car to a stop in about 2 inc

8_murik_8 [283]

Explanation:

Using equation of motion to determine the acceleration of the car,

vf^2 = vi^2 + 2 * a * S,

vf = 0

0 = vi^2 + 2 * a * S

Converting mph to m/s,

3 mph * 5280 ft/mi * 12 in/ft * 2.54 cm/in * 1 m/100 cm * 1 h/3600 s

= 3 * 0.445

v = 1.335 m/s

Converting in to m,

2 in * 2.54 cm/in * 1 m/100 cm = 0.0254 m

= 2 * 0.0254

S = 0.0508 m

0 = 1.335^2 + 2 * a * 0.0508

a = -1.335^2 ÷ 0.1013

= -17.54 m/s^2

Mass of car (assumed) = 2000 kg

Force = ma

= 2000 × 17.54

= 35.08 kN.

5 0

4 years ago