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Aleks04 [339]
3 years ago
12

Un cuerpo se mueve con MRU a una rapidez de 12 m/s y recorre 240 m; luego, disminuye su ra¬pidez uniformemente a razón de 2 m/s2

hasta que adquiere una rapidez de 2 m/s2, en ese momento acelerar uniformemente a razón de 5 m/s2 hasta adquirir una rapidez de 22 m/s. Calcula el tiempo que empleó hasta dicho momento.
Physics
1 answer:
Annette [7]3 years ago
7 0

Answer:

29 seconds

Explanation:

First we have a constant speed of 12 m/s and the distance of 240 m, so to find the time we can use the formula:

distance = speed * time

240 = 12 * time1

time1 = 20 seconds

Then, the speed decreases at 2 m/s2 until it reaches 2 m/s. So to find this time, we use this formula:

Final speed = inicial speed + acceleration * time

2 = 12 - 2 * time2

2*time2 = 10

time2 = 5 seconds.

Then, the speed increases from 2 m/s to 22 m/s with an acceleration of 5 m/s2, so we have:

Final speed = inicial speed + acceleration * time

22 = 2 + 5 * time3

5*time3= 20

time3 = 4 seconds

The total time is:

Total time = time1 + time2 + time3 = 20 + 5 + 4 = 29 seconds

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A wheelbarrow is a good example of a second-class lever. True or False
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What is the direction of the force for a negative charge moving downward in a magnetic field pointing to the left?
velikii [3]

Answer: A

Out of the screen

Explanation:

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What type of animals does Dr. Grant study?
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7 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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