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3 years ago

11

Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi

s process?

2 answers:

morpeh [17]3 years ago

4 0

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = $\dfrac{0.029158}{4.03176}$ = 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

Xelga [282]3 years ago

4 0

Answer:

(a) 992.87 g

(b) $6.419\times 10^{14} J$

Solution:

As per the question:

Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g

(a) To calculate mass of He produced:

We know that:

Atomic mass of hydrogen is 1.00784 u

Also,

4 Hydrogen atoms constitutes 1 Helium atom

Mass of Helium formed after conversion:

$4\times 1.00784 = 4.03136 u$

Also, we know that:

Atomic mass of Helium is 4.002602 u

The loss of mass during conversion is:

4.03136 - 4.002602 = 0.028758 u

Now,

Fraction of lost mass, M' = $\frac{0.028758}{4.03136} = 0.007133 u$

Now,

For the loss of mass of 1000g = $0.007133\times 1000$ = 7.133 g

Mass of He produced in the process:

$M_{He} = 1000 - 7.133 = 992.87 g$

(b) To calculate the amount of energy released:

We use Eintein' relation of mass-enegy equivalence:

$E = M'c^{2}$

$E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J$

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The correct answer is A.

The cell membrane consists of a phospholipid bilayer with embedded proteins. Sometimes molecules are just too big to easily flow across the plasma membrane or dissolve in the water so that they can be filtered through the cell membrane. In these cases , the cells must put out a little energy to help get molecules in and out of the cell.

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4 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

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