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3 years ago

5

Which of the following does NOT influence one's body composition? A. metabolism B. financial problems C. childhood obesity D. me

dical problems Please select the best answer from the choices provided.

2 answers:

dlinn [17]3 years ago

6 0

Option B i.e Financial Problems is the correct answer.

Metabolism (Conversion of food into energy), childhood obesity (over-weight) and medical problems (disease or other health problem) are directly related to child's health and thus influences his body composition. On the other hand, a child does not have any financial problems or financial pressure and thus does not influence his body composition.

nikklg [1K]3 years ago

4 0

I think it is B because i has nothing to do with body comp.

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe

telo118 [61]

Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

Given

Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

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2 years ago

If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?

Scorpion4ik [409]

Answer:

2+1

Explanation:

2+1

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2 years ago

Read 2 more answers

Which is the electric potential energy of a charged particle divided by its charge?Electric fieldelectric field lineelectric pot

Dahasolnce [82]

<h3><u>Answer;</u></h3>

electric potential

<h3><u>Explanation;</u></h3>

Electric potential is the electric potential energy per unit charge.

Mathematically; V =PE/q

Where; PE is the electric potential energy, V is the electric potential and q is the charge.

Electric potential is more commonly known as voltage. If you know the potential at a point, and you then place a charge at that point, the potential energy associated with that charge in that potential is simply the charge multiplied by the potential.

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3 years ago

Help ASAP <br> Just answer the first question for me please!

Usimov [2.4K]

Answer:

Im pretty sure its b

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