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3 years ago

A good model of the cell membrane would be

Physics

1 answer:

Nonamiya [84]3 years ago

6 0

B. A film of food wrap. The cell membrane is the outermost layer that is protecting the cell

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How many significant digits are in the following measurements?<br> a. 1300 m

Fofino [41]

Answer:

For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).

Explanation:

hope it helps :)

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2 years ago

What do you have to do to become faster at running?

GenaCL600 [577]

Answer: Get stronger. Strength is critical to running fast because the more force you can exert against the ground the faster you can run.

Apply your new strength. Being strong is important for sprinting, but the ability to apply it is even more important. Sprinters move in a horizontal direction.

Take longer steps more quickly. Today, the concept of exerting more force against the ground is very popular for improving speed. ...

Run fast to get fast. Of all the strategies described in this article, this is the most important and often the most overlooked.

Explanation:

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3 years ago

Read 2 more answers

You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The

AnnZ [28]

Answer:

0.75

Explanation:

Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.

Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).

So, μ = F/N

= 300 N/400 N

= 3/4

= 0.75

So, the coefficient of static friction μ = 0.75

6 0

3 years ago

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the

GarryVolchara [31]

Explanation:

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).

For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

$P^2=kA^3$

k is constant of proportionality

It is required to solve the above equation for k

$k=\dfrac{P^2}{A^3}$

8 0

3 years ago

B. Projectile on cliff (range)

dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

cos 25 = v₀ₓ / v₀

sin 25 = Iv_{oy} / v₀

v₀ₓ = v₀ cos 25

v_{oy} = v₀ sin 25

v₀ₓ = 22 cos 25 = 19.94 m / s

v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

0 = 21 + 0.0192 t - ½ 9.81 t²

4.905 t² - 0.0192 t - 21 = 0

t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

t = $\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}$

t = $\frac{0.003914 \ \pm 4.13828}{2}$

t₁ = 2.07 s

t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

t = 2.07 s

now we can look up the distance traveled

x = v₀ₓ t

x = 19.94 2.07

x = 41.28 m

5 0

3 years ago