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2 years ago

What body process converts physical energy to electrical energy?

Physics

1 answer:

nevsk [136]2 years ago

7 0

The correct answer of the question given is D.

There are four body processes mentioned in the question are named as glycolysis,metabolism,photosynthesis and transduction.

Glycolysis is the metabolic process in which glucose is converted to pyruvic acid with the release of ATP and NADPH.

Metabolism is a chemical process which includes the formation or dissociation of chemical compounds.

Photosynthesis is the biological process in which food is prepared by plants by absorbing minerals, water from soil and carbon dioxide from the atmosphere in the presence of radiant energy [sun light] .

Transduction: It is the biological process in which any external physical stimuli is received and relayed from one part of the body to another part.During this process energy i.e stimuli is converted into electro chemical stimuli.

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An electron is trapped in a square well of unknown width, L. It starts in unknown energy level, n. When it falls to level n-1 it

Lady bird [3.3K]

Answer:

(1) En to n-1 = 0.55 ev

(2) En-1 to n-2 = 0.389 ev

(3) ninitial =4

(4) L =483.676 ×10^-11 nm

(5) λlongest= 1773.33 nm

Explanation:

Detailed explanation of answer is given in the attached files.

4 0

2 years ago

a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr

mash [69]

Answer:

the inductive reactance of the coil is 1335.35 Ω

Explanation:

Given;

inductance of the coil, L = 250 mH = 0.25 H

effective current through the coil, I = 5 mA

frequency of the coil, f = 850 Hz

The inductive reactance of the coil is calculated as;

$X_l = \omega L = 2\pi f L\\\\X_l = 2\pi \times 850 \times 0.25\\\\X_l = 1335.35 \ ohms$

Therefore, the inductive reactance of the coil is 1335.35 Ω

6 0

1 year ago

A sleeping 68 kg man has a metabolic power of 79 w .

Lesechka [4]

3 0

2 years ago

Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?

Nataly [62]

ANSWER

$\theta=35\degree$

EXPLANATION

Since the body is in equilibrium, total upward forces must equal total downward force.

Also the net horizontal forces acting on the body must be zero.

We need to resolve $F_1$ into vertical and horizontal components.

The horizontal component is

$x=F_1\cos\theta$ .

The vertical component is

$y=F_1\sin\theta$ .

Equating the up force to the downward forces gives,

$F_1\sin\theta + 20N=60N$ .

This implies that,

$F_1\sin\theta =60N-20N$ .

$F_1\sin\theta=40N...eqn1$

Also the horizontal forces must be equal.

$F_1\cos\theta=57N...eqn2$ .

Dividing equation (1) by equation (2) gives,

$\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}$ .

$\Rightarrow \tan\theta=0.70175$

$\Rightarrow \theta=tan^{-1}(0.70175)$

$\Rightarrow \theta=35.0594$ .

Therefore the given angle that $F_1$ must make with the horizontal is approximately 35° to the nearest degree.

4 0

2 years ago

Read 2 more answers

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

2 years ago