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ira [324]
3 years ago
12

Pls help me with this question its due tomorrow and I can't find the answer: A person jumped 50cm, the hang time for the jump wa

s 0.778 seconds, calculate a value for the acceleration due to gravity. What is the percent error in the value for acceleration you found? (Solve it without factoring in the air)

Physics
1 answer:
son4ous [18]3 years ago
4 0

S = ut + 1/2gt²

where S = distance

50 cm = 0.5m

u is initial velocity = 0

S = 0*t + 1/2gt²

S = 1/2gt²

0.5 = 1/2 g (0.778²)

2*0.5 = g(0.778²)

1/(0.778²) = g

g = 1.65m/s²

The value of g is grossly errored due to the values obtained since the known value of g on earth is 9.8m/s² the obtained value rounds off to the value of g on moon.

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
3 years ago
In order for a living organism to substain its life it must be able to
schepotkina [342]
Have the 6 characteristics of living things
1.movement/growth
2.reproduction
3.made of 1 or more cells
4.sensitivity
5.excretion
6.nutrition
8 0
3 years ago
You have decided to study the effect of loud noise on plant growth. You put one plant in a quiet room and the other plant in you
wel
The answer is B. Controlled variable.<span />
4 0
3 years ago
Read 2 more answers
Metals are good conductors of electric current for which of the following reasons?They possess high concentrations of protonsThe
KatRina [158]

The correct answer is: They possess high concentrations of free electrons

The main characteristic of good conductors such as metals is the presence of movable electrically charged particles, or electrons. So, when an electric current is applied to a metal, the electrons will move and allow electricity to pass through them. Materials opposite of metals, with low electron mobility are not good conductors, instead they are called insulators.

3 0
3 years ago
Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha
Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

   The correct option is b

Explanation:

From the question we are told that

          The charge carried by A is  q C

           The charge carried by B is 0 C

            The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by  A and B is mathematically evaluated as

                 \frac{ 0 + q}{2} =   \frac{q}{2}

When C and B are brought close and then separated the charge carried by  C and B  is mathematically evaluated as    

                    \frac{0 + \frac{q}{2} }{2}  = \frac{q}{4}

When C and A are brought close and then separated the charge carried by  C and A  is mathematically evaluated as  

                       \frac{\frac{q}{4} +  \frac{q}{2} }{2}   = \frac{3q}{8}

Looking at these calculation we can see that the charge carried by B is

        \frac{q}{4} C

8 0
3 years ago
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