Where would you find the lowest-density seawater?

What did the moon of Titan contain that is of interest to scientists?

statuscvo [17]

The answer to this question is C. Lakes of hydrocarbons. Trust me this is the answer before i got the results i thought it was lakes of mercury.

7 0

3 years ago

Read 2 more answers

The light reactions could be viewed as analogous to a hydro-electric dam. In that case, the wall of the dam that holds back the

Viktor [21]

The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.

In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.

If you need to learn more about light reactions click here:

brainly.com/question/26623807

#SPJ4

3 0

1 year ago

A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve

Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

$s_2 = gt^2/2$

$s_1 = g(t + \Delta t)^2/2$

$v_2 = gt$

$v_1 = g(t + \Delta t)$

Their difference in distance are therefore:

$\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2$

$\Delta s = g/2((t + \Delta t)^2 - t^2)$

$\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t)$ (As $A^2 - B^2 = (A-B)(A+B)$

$\Delta s = g\Delta t/2(2t + \Delta t)$

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

$\Delta v = v_1 - v_2 = g(t + \Delta t) - gt$

$\Delta v = gt + g\Delta t - gt = g\Delta t$

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0

3 years ago

Read 2 more answers

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$