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3 years ago

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

Physics

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:

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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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Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete

harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

$v=\dfrac{d}{T}$

d is distance, $d=2\pi r$ and 1/T = f (frequency)

$v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s$

So, the average speed of the rubber stopper is 18.84 m/s.

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3 years ago

Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th

tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

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2 years ago

To shoot a swimming fish when an intense light beam from a laser gun you should aim

Ivan

Answer

aim directly at the image

Explanation

the light from the laser beam will also bend when it hits the air water interface , so aim directly at the fish

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3 years ago

___ + 3H2O + light —> C3H6O3 + 3O2. What amount and substance balance this reaction?

Alik [6]

Answer:3H2O + light-c3h603+302

Explanation:

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3 years ago

Read 2 more answers

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

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3 years ago