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lara [203]
3 years ago
13

During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma

ss of the black residue formed was 79.6 g. Does the law of conservation of mass hold true in this case? Use complete sentences to justify your answer based on numerical calculations.
Physics
1 answer:
iris [78.8K]3 years ago
3 0

When copper carbonate is heated, it decomposes to form a black residue. This black residue is actually copper(II) oxide. Along with this black residue, carbon-dioxide is released. The complete reaction is the following:

CuCO_3\to CuO(s)+CO_2(g)

The law of conservation of mass holds true. 123.6 g of copper carbonate decomposes to 79.6 g of copper oxide and (123.6 g-79.6 g=44 g) 44 g of carbon-dioxide.

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Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      E_A} =E_{B}

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg(h_{B}-h_{A} )=mgh

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                                 k"=1.13 *k= 1.13*9568.92

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5 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
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1) 1.86\cdot 10^6 rad/s^2

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4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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Answer:

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I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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