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3 years ago

Why does 30 kg mass at the top of a hill have more potential energy?

1 answer:

6 0

So when it comes to Gravitational Potential Energy the higher an object of mass is the more energy it has as the equation for EPG is = MGH so the M is Mass of the object , the G is the gravitational constant which on earth is roughly 9.8 m/s and the H is the height of the object. So the greater the H value is the more energy you'll have. An example would be using your 30kg mass say at a height of 10 meters would have the EPG of 2940 Joules of energy since EPG= MGH so EPG= (30)(9.8)(10) . Now if the object was at a height of 20 meters the EPG would be greater as EPG=(30)(9.8)(20) which would be 5880 joules of energy

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Technician A says some compressor service procedures can be performed on the vehicle if space permits. Technician B says modern

stiv31 [10]

Answer:

Technicians A.

Explanation:

Since air compressor uses series of processes that turn incoming ambient air into a power source for tools and machinery. This means that air compressor has many different parts, and each of these parts must be maintained to ensure they function properly and optimally.

These are the basis when it comes to servicing a compressor

You need to change its oil

And clean its filters.

Inspected it's filters every three months, and have its filters replaced and connections tightened at least once every year.

To do all these can be performed on the vehicle if there is enough space just as Technician A said for the question context.

5 0

3 years ago

Luis wanted to know which of his 4 toy cars is the fastest. He conducted an experime and recorded his results in the table shown

Murrr4er [49]

The dependent variable was the time and the independent variable was thecars

6 0

3 years ago

a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition

MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0

2 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0

3 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)

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

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

*****

4 0

3 years ago