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leva [86]
3 years ago
14

What part of the phospholipid bilayer “hates” water?

Physics
2 answers:
Juliette [100K]3 years ago
5 0

Answer:

Phospholipids are able to form cell membranes because the phosphate group head is hydrophilic (water-loving) while the fatty acid tails are hydrophobic (water-hating)

VladimirAG [237]3 years ago
3 0

Answer:

Explanation:

the fatty acid tails are hydrophobic

they hates water

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What is the distance write the equation
Sedbober [7]

Answer:

135 m

Explanation:

Given:

vi = 12 m/s

vf = 18 m/s

t = 9s

Find: x

x = ½ (vf + vi) t

x = ½ (18 m/s + 12 m/s) (9 s)

x = 135 m

5 0
3 years ago
A stone is dropped off a cliff and it falls for 4.8 m/s
valina [46]

Answer:

you cant caculate this because there is no question

Explanation:

7 0
3 years ago
- A cannon of 2000 kg fires a shell of 10 kg at
fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

p=0

After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

0=MV+mv

And solving for V, we find the velocity of the cannon:

V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

7 0
3 years ago
A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler
ki77a [65]

Answer:

a)t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec

b)\theta_1=\frac{w_1^2}{2\alpha}rad

c)t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

w=\frac{\Delat \theta}{\Delta t}

Angular acceleration is the rate of change of the angular velocity respect to the time

\alpha=\frac{dw}{dt}

2) Part a

We can define some notation

w_o=0\frac{rad}{s},represent the initial angular velocity of the wheel

w_1=?\frac{rad}{s}, represent the final angular velocity of the wheel

\alpha, represent the angular acceleration of the flywheel

t_1 time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

w_1=w_o +\alpha t_1

And solving for t1 we got:

t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

\Delta \theta=wt+\frac{1}{2}\alpha t^2

Replacing the values for our case we got:

\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2

And we can replace t_1from the result for part a, like this:

\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2

Since \theta_o=0 and w_o=0 then we have:

\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}

And simplifying:

\theta_1=\frac{w_1^2}{2\alpha}rad

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is \alpha_1 =-5\alpha \frac{rad}{s}

We have an initial angular velocity w_1, and since at the end stops we have that w_2 =0

Assuming that t_2 represent the time in order to stop the wheel, we cna use the following formula

w_2 =w_1 +\alpha_1 t_2

Since w_2=0 if we solve for t_2 we got

t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}

And from part a) we can see that w_1=\alpha t_1, and replacing into the last equation we got:

t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec

5 0
3 years ago
Need help with this! <br><br> just scienceeeeeee
LiRa [457]

Answer:

7)  λ = 0.5 m,  8)  f = 4.8 10¹⁴ Hz

Explanation:

The speed of an electromagnetic wave is

          c = λ f

where c is the speed of light in vacuum c = 3 10⁸ m / s

7) indicate the frequency f = 6.0 10⁸ Hz

we do not know the wavelength

         λ = c / f

       

we calculate

        λ = 3 10⁸ / 6.0 10⁸

        λ = 0.5 m

8) indicate the wavelength  λ = 6.25 10-7 m

we do not know the frequency

         f = c / λ

we calculate

        f = 3 10⁸ / 6.25 10⁻⁷

        f = 0.48 10¹⁵ Hz

        f = 4.8 10¹⁴ Hz

8 0
3 years ago
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