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s344n2d4d5 [400]
3 years ago
15

Please explain in detail why it is possible to estimate, to a good approximation, the energy derived from a sugar cube in the bo

dy, by doing laboratory experiments. Obviously the mechanisms are completely different and your answer should address this issue.
Physics
1 answer:
Kaylis [27]3 years ago
8 0
The reactions that are carried out on sugar within are body are very similar to the reaction of combustion. The reactants used and the product formed are the same in the aerobic respiration reaction. Therefore, it is possible to estimate the energy released by the sugar in the body by finding out the energy released when the sugar is burned. Despite the products arising from different pathways with different mechanisms, the reactants and products being the same means that the energy released will be similar.
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Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950
Serjik [45]

Answer:

this description is valid for mediadle displacement, bone is an acceptable description

Explanation:

The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.

In the description this has a starting point corner NO of pine and 675.

Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.

After analyzing  this description is valid for mediadle displacement, bone is an acceptable description

6 0
3 years ago
What is potential energy?
iVinArrow [24]
energy associate with position or shape
5 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
6. Mr. Leppold jumps out of a plane with a parachute...before the chute opens,
polet [3.4K]

1) He has both potential and kinetic energy

2) Before the parachute opens, the potential energy decreases and the kinetic energy increases

Explanation:

1)

The gravitational potential energy of a body is the energy possessed by the object due to its position in a gravitational field, and it is given by:

PE=mgh

where

m is the mass of the body

g is the acceleration of gravity

h is the height of the body above the ground

On the other hand, the kinetic energy of a body is the energy possessed by the body due to its motion; it is given by

KE=\frac{1}{2}mv^2

where

v is the speed of the object

Here Mr. Leppold has both potential and kinetic energy before opening the parachute, because:

- It is moving at a certain speed, so v\neq 0, therefore he has kinetic energy

- He is at a certain height above the ground, h\neq 0, therefore he has potential energy

2)

The total mechanical energy of Mr.Leppold is the sum of the potential and the kinetic energy:

E=PE+KE

According to the law of conservation of energy, in absence of air resistance, this quantity remains constant.

During the fall, the height of Leppold decreases: this means that as h decreases, the potential energy decreases  too.

However, the total energy E must remain constant: therefore, this means that the kinetic energy KE must increase, and this occurs because the speed of Mr. Leppold increases as he falls.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0
3 years ago
A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the
tatiyna

Answer:

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁)................... Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t................... Equation 2

make t  the subject of the equation

t = Q/P................. Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.

6 0
3 years ago
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