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1 year ago

A large tank is filled with water to a depth of 38 m. An opening is located 14 m

1 answer:

5 0

When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank then the water would emerge from the opening at a velocity of 21.70 m/s

<h3>What is mechanical energy?</h3>

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.

As total mechanical energy ias s the sum of all the kinetic as well potential energy stored in the system.

ME = KE + PE

The total kinetic energy

KE = 1/2*m*v²

Similarly, the total potential energy

PE = m*g*h

As given in the problem When a large tank is filled with water to a depth of 38 m. An opening is located 14 m above the bottom of the tank

height of the water above the opening

h= 38-14

h= 24 m

The potential energy of water above the opening is getting converted into the kinetic energy

m*g*h = 1/2*m*v²

v = √(2*g*h)

By substituting the respective values of the height and gravity

v = √(2*9.81*24)

v = 21.70 m/s

Thus, when water is poured into a sizable tank until it reaches a 38 m depth. If a hole were to be made at a height of 14 m above the tank's bottom, water would rush out of it at a speed of 21.70 m/s.

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2 years ago

A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop

Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward = 2.8m

Time of travel = 5min

Distance walked eastward = 9.2m

Time of travel = 10min

Unknown:

The total shopper's travel distance = ?

Solution:

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How do solar panels work with conduction, convection and radiation?

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Explanation:

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3 years ago

object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Cal

VARVARA [1.3K]

The distance between object P1 and its image formed is determined as 36 m.

<h3>Distance of the image</h3>

The distance of the image formed by object P1 is calculated as follows;

In a plane mirror; object distance = image distance

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distance between object and image = 18m + 18 m = 36 m

Thus, the distance between object P1 and its image formed is determined as 36 m.

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2 years ago

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

$W= F*d= m*g*d$

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

$W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]$

Converting from Joules to Kcals:

$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$

$P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]$

*****

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3 years ago