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18 days ago

In a fusion reaction, the nuclei of two atoms join to form a single atom of a different element. In such a reaction, a fraction

of the rest energy of the original atoms is converted to kinetic energy of the reaction products. A fusion reaction that occurs in the Sun converts hydrogen to helium. Since electrons are not involved in the reaction, we focus on the nuclei. Hydrogen and deuterium (heavy hydrogen) can react to form helium plus a high-energy photon called a gamma ray: $^1 H + ^2 H \longrightarrow ^3 He + \gamma$ Objects involved in the reaction:

Physics

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A kinesin that is transporting a secretory vesicle uses approximately 80 ATP molecules/s. Each ATP provides a kinesin molecule w

inna [3103]

Answer:

The force is $F = 8*10^{-12} \ N$

Explanation:

According to the inquiry, we understand that

The rate at which ATP molecules are utilized is $R = 80 ATP/ s$

The energy yielded by a single ATP molecule is $E_{ATP} = 0.8 * 10^{-19} J$

The kinesin's velocity is $v = 800 nm/s = 800*10^{-9} m/s$

The power generated by the ATP in one second can be expressed mathematically as

$P = E_{ATP} * R$

After substituting the values

$P = 80 * 0.8*10^{-19 }$

$P = 6.4 *10^{-18}J/s$

Now this power can be represented mathematically as

$P = F * v$

Where F indicates the force exerted by the kinesin

Therefore

$F = \frac{P}{v}$

after substituting input values

$F = \frac{6.4*0^{-18}}{800 *10^{-9}}$

$F = 8*10^{-12} \ N$

7 0

3 months ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

kicyunya [3294]

Answer:

Estimate of the sample's volume: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Mean density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
The volume of the cord is considered negligible.

Explanation:

Overall volume of the sample

The magnitude of the buoyant force equals $\rm 17.50 - 11.20 = 6.30\; N$ .

This also corresponds to the weight (weight, $m \cdot g$ ) of the water displaced by the object. To determine the mass of the displaced water from its weight, apply the formula: divide weight by $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assuming the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To find the volume of the displaced water, use the formula: divide mass by density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assuming the cord's volume is negligible, since the sample is completely submerged in water, its volume should equal the volume of the displaced water.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

Mean Density of the sample

Average density can be calculated by the mass divided by volume.

To compute the mass of the sample from its weight, utilize the formula: divide by $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume from the previous section can be utilized.

Lastly, divide mass by volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

2 months ago

An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time

Keith_Richards [3271]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

3 0

4 months ago