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3 years ago

8

Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released):

2 answers:

algol133 years ago

5 0

Answer:

The correct answer is

Endothermic:

AB+C-->AC+B

A2+C2-->2AC

B2+C2-->2BC

Exothermic:

A+BC-->AB+C

A2+B2-->2AB

Explanation:

Trust Me

VashaNatasha [74]3 years ago

4 0

Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released):
Endothermic
B2+C2-->2BC
A2+C2-->2AC
<span>A2+B2-->2AB</span>

Exothermic
AB+C-->AC+B
A+BC-->AB+C

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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3 years ago

Where do we hear loud sound, at antinode or node?

tester [92]

Answer:

node

Explanation:

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Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

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$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

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To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

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If h is the height of the tower, the energy on top of the tower:

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Combining equation 2 and 3 and solving for h:

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Answer:

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