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TEA [102]
3 years ago
13

HELP ASAPP!!!! WILL GIVE BRAINLIEST! Describe the chemical structure of a cell membrane

Physics
2 answers:
MariettaO [177]3 years ago
4 0

Answer:

The plasma membrane is composed of a bilayer of phospholipids, with their hydrophobic, fatty acid tails in contact with each other. ... Carbohydrates are attached to some of the proteins and lipids on the outward-facing surface of the membrane. These form complexes that function to identify the cell to other cells.

Explanation:

Hope it answered you plz mark as Brainlist

bagirrra123 [75]3 years ago
3 0
Main components are lipids, proteins, and carbohydrates
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Help with Order of Magnitude question?
solong [7]
The rotation of Earth is equivalent to one day which is comprised of 24 hours. To determine the number of miles in Earth's circumference, one simply have to multiply the given rate by the appropriate conversion factor and dimensional analysis. This is shown below.
 
                  C = (1038 mi/h)(24 h/1 day)
                   C = 24,912 miles

From the given choices, the nearest value would have to be 20,000 mile. The answer is the second choice. 
7 0
3 years ago
An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h

Now,

Average speed = total distance/total time

V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}

Hence, the average speed for the trip is 270 mi/h.

3 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
A car, starting from rest, accelerates at
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Using V= vo +at with Vo = 0 and a= 4m/sec2.
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3 0
3 years ago
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Kaylis [27]
It can be a) 12Hz.................
5 0
3 years ago
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