The answer is not B.

How might the velocity (speed) of wind or water affect the deposition of sediments?

Maksim231197 [3]

Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
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6 0

3 years ago

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

8 0

3 years ago

Refraction can occur only when light is

mr_godi [17]

Answer:

We have learned that refraction occurs as light passes across the boundary between two media. Refraction is merely one of several possible boundary behaviors by which a light wave could behave when it encounters a new medium or an obstacle in its path.

3 0

2 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

Soloha48 [4]

Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

4 0

3 years ago

A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory (ignorin

Nat2105 [25]

Answer:

b) Gravity

Explanation:

Gravity acts all of the time, when you apply force to a projectile it has to be more than the forces of the gravity and air resistance together so the projectile can move, when the rock is at the top of its trajectory the force that you applied at the beginning is getting lost, so the other forces (air resistance and gravity) make the rock fall to the floor.

8 0

3 years ago