Ask question

Ask question

All categories

2 days ago

7

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

ith an energy of about 0.8 × 10-19 J. If the velocity of the kinesin is 800 nm/s, can you determine the force the kinesin is exerting, if you assume that all the ATP energy is used (100% efficiency)? If you can, find it and give your answer in newtons. If not, answer with 0.

1 answer:

inna [987]2 days ago

7 0

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

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

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

You might be interested in

Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average

ValentinkaMS [1144]

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

7 0

1 day ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [1153]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

17 days ago

A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz

serg [1189]

Answer

Data provided:

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

distance of slide = 10 m

height of the block = 4 m

calculation for the block's potential energy

P = m g h

P = 0.2 x 9.8 x 4

P = 7.84 J

kinetic energy calculated as

$KE = \dfrac{1}{2}mv^2$

$KE = \dfrac{1}{2}\times 0.2 \times 7.84^2$

$KE =6.14 J$

Work done = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) using the formula v² = u² + 2 a s

0 = 8² - 2 x a x 10

a = 3.2 m/s²

ma - μ mg = 0

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{3.2}{9.8}$

$\mu = 0.327$

7 0

8 days ago

Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

serg [1189]

Answer:

b) TA = TB = TC

Explanation:

When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
The equilibrium state will be established once this equation is satisfied:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
Therefore, option b) is correct.

8 0

5 days ago

A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in

Maru [1053]

Answer:

Explanation:

The transitions occur as follows:

P₁ V₁ changes to 3P₁, V₁ (with constant volume) — first phase.

Subsequently, 3P₁,V₁ transitions to 3P₁, 5V₁ (with constant pressure) — second phase.

During the initial phase, the temperature must be escalated by a factor of 3. Therefore, if the starting temperature is T₁, then the ending temperature will be 3 T₁.

P₁V₁ = n R T₁, where n represents the number of moles of gas.

Thus, nRT₁ = P₁V₁.

The heat added at constant volume is given by n Cv (3T₁ - T₁),

= n x 5/3 R X 2T₁ (noting that for diatomic gas, Cv = 5/3 R).

= 10/3 x nRT₁

= 10/3 x P₁V₁.

In the second phase, the temperature must rise 5 times. Thus, if the initial temperature is 3T₁, then the final temperature will be 15 T₁.

The heat added at constant pressure in this scenario becomes:

= n Cp (15T₁ - 3T₁)

= n x 7/3 R X 12T₁ (for diatomic gases, Cp = 7/3 R).

= 28 x nRT₁

= 28 P₁V₁.

6 0

12 days ago