All categories

3 years ago

What is the cell structure made of a phospholipid layer?

Physics

2 answers:

Dominik [7]3 years ago

6 0

Answer:

Cell Membrane

Explanation:

The cell membrane contains a phospholipid bilayer.

jarptica [38.1K]3 years ago

4 0

The answer is cell membrane.

You might be interested in

GIVING BRAINLIEST PLEASE HELP!!

OLga [1]

A. because I had this question yesterday.

3 0

3 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

3 years ago

The strength of the electric field of a charged particle becomes greater as the distance from the particle increases. T F

natali 33 [55]

The answer is T it is true

7 0

3 years ago

6. A 73-kg woman stands on a scale in an elevator. The

olga nikolaevna [1]

Answer:

Explanation:

n = m(g +a)

n= normal force (N)

m=mass (kg)

g=acceleration of gravity

a= acceleration of elevator

rearrange:

a= n/m - g

a= (810 N/73 kg) - 9.8 m/s ^2

a= 1.3 m/s ^2 up

and the acceleration is upwards bc her weight is less than the scale reading

3 0

3 years ago

A basketball player spins the ball with an angular acceleration of 10 rad/s2. What is the ball’s final angular velocity if the b

Lena [83]

Explanation:

It is given that,

The angular acceleration of the basketball, $\alpha=10\ rad/s^2$

Time taken, t = 3 seconds

We need to find the ball’s final angular velocity if the ball starts from rest. It can be calculated using definition of angular acceleration i.e.

$\alpha=\dfrac{\omega_f-\omega_i}{t}$

$\omega_i=0\ (rest)$

$\omega_f=\alpha t$

$\omega_f=10\ rad/s^2\times 3\ s$

$\omega=30\ rad/s$

So, the ball's final angular velocity is 30 rad/s. Hence, this is the required solution.

8 0

3 years ago