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3 years ago

Which term refers to the diffusion of water through a membrane?

Physics

2 answers:

MrRa [10]3 years ago

7 0

A.) osmosis. Ahhh-smosis and ahhh-qua (water in Spanish).

marin [14]3 years ago

4 0

A. Osmosis would be the answer to the diffusion of water through a membrane.

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Which type of wave affects the surface of the land by causing it to rise and fall like waves on an ocean?

denis-greek [22]

It is a seismic wave that effects the surface of the land by causing it to rise and fall like waves in an ocean. Seismic waves can be broken down into a number of different types. Specifically, it is a Raleigh surface wave that behaves much like water waves since it moves the ground up and down as it travels. Seismic waves are produced in earthquakes.

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3 years ago

6 jam jars of volume 0.37m3 are filled with Jam. An empty jam jar has a weight 200g each. The weight of all 6 jars filled with j

marta [7]

Answer:

0.14594 g/cm³

Explanation:

Density = Mass / Volume

Mass = 5400 g

Volume = 0.37 m³

Density = M /V

volume = 0.37 m³ = 37000 cm³

Density = $\frac{5400}{37000}$

= 0.14594 g/cm³

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3 years ago

I drove from Oklahoma to California (1452.9 miles) in 23 hours? What was my average speed?

storchak [24]

Answer:

63 miles per hour

Explanation:

you just divide miles and hours to get your average speed

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3 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0

2 years ago

Read 2 more answers

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0

4 years ago

Read 2 more answers