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Paha777 [63]
3 years ago
9

Which of the following are the advantages of a DBMS (database management system)?

Physics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer:

A

Explanation:

You might be interested in
How does the time at which you see things happen at a baseball game compare to the time when you hear things happen? Explain you
hoa [83]

Answer:

Because of the speed of the sound.

Explanation:

The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].

Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.

With the following equation, we can calculate how long it takes to hear a hit or explosion

t = x / v

where:

x = distance [m]

v = sound velocity = 345 [m/s]

t = time [s]

7 0
3 years ago
during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f
vovangra [49]

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

3 0
2 years ago
Explain in terms of energy flow how a cold pack works on a sprained ankle
sergij07 [2.7K]
The <span>flow of how a cold pack works on a sprained ankle is based on the second law of thermodynamics which states that energy will flow from a higher to a lower temperature. So your body heat will flow to the cold pack in which you will feel the coldness of the pack.</span>
7 0
2 years ago
A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the
Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

- We will use differential calculus to solve this problem.

- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

                                             F = 78480*x*dx

- Now, the work done is given by:

                                             W = F.s

- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

- We have the work as follows:

                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

5 0
3 years ago
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