All categories

3 years ago

HI I am taking a quiz and I need help it’s science o.o:

Physics

1 answer:

MissTica3 years ago

5 0

Answer:

it's true, I think.

Explanation:

A cell's membrane is primarily made up of a double layer of phospholipids (fatlike, phosphorus-containing substances). Each layer is composed of phospholipid molecules that contain a hydrophilic (water-loving) head and a hydrophobic (water-repellent) tail.

have a wonderful day! :)

You might be interested in

The Golden Gate Bridge in San Francisco has a main span of length 1.28 km, one of the longest in the world. Imagine that a steel

Valentin [98]

Answer:

361.46 N

Explanation:

$\alpha$ = Coefficient of thermal expansion = $11\times 10^{-6}\ /^{\circ}C$

Y = Young's modulus for steel = $20\times 10^{10}\ Pa$

A = Area = $3.1\times 10^{-6}\ m^2$

$L_0$ = Original length = 1.28 km

$\Delta T$ = Change in temperature = $45-(-10)$

Length contraction is given by

$\Delta L=\alpha L_0\Delta T$

Also,

$\Delta L=\dfrac{L_0T}{YA}$

$\alpha L_0\Delta T=\dfrac{L_0T}{YA}\\\Rightarrow T=\alpha \Delta TYA\\\Rightarrow T=11\times 10^{-6}\times (43-(-10))\times 20\times 10^{10} \times 3.1\times 10^{-6}\\\Rightarrow T=361.46\ N$

The tension in the wire is 361.46 N

4 0

3 years ago

A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s

Komok [63]

1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
$2aS = v_f^2-v_i^2 = -v_i^2$
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
$a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2$

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
$F=ma$
This force will be constant, and since m is always the same, then a is the same even in the second situation.

3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
$2aS=-v_i^2$
$S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m$

4 0

3 years ago

How much money can be found in the pockets/purses of the 250 students in a class if no student may carry 2 identical banknotes w

vaieri [72.5K]

The minimum is $1- $5000000

5 0

3 years ago

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h

Aneli [31]

Answer:

On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let $k$ denote the Coulomb constant, and let $q$ denote the size of a point charge. At a distance of $r$ away from the charge, the electric field due to this point charge will be:

$\displaystyle E = \frac{k\, q}{r^2}$ .

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let $q_1 = 5\times 10^{-19}\; \rm C$ and $q_2 = 20 \times 10^{-19}\; \rm C$ . Assume that the electric field is zero at $r$ meters to the right of the $5\times 10^{-19}\; \rm C$ point charge. That would be $(2 - r)$ meters to the left of the $20 \times 10^{-19}\; \rm C$ point charge. (Since this point should be between the two point charges, $0 < r < 2$ .)

The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}$ .

The electric field due to $q_2 = 20 \times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}$ .

Note that at all point in this section, the two electric fields $E_1$ and $E_2$ will be acting in opposite directions. At the point where the two electric fields balance each other precisely, $| E_1 | = | E_2 |$ . That's where the actual electric field is zero.

$| E_1 | = | E_2 |$ means that $\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}$ .

Simplify this expression and solve for $r$ :

$\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0$ .

$\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0$ .

Either $r = -2$ or $\displaystyle r = \frac{2}{3}\approx 0.67$ will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, $0 < r < 2$ . Therefore, $(-2)$ isn't a valid value for $r$ in this context.

As a result, the electric field is zero at the point approximately $0.67\; \rm m$ away the $5\times 10^{-19}\; \rm C$ charge, and approximately $2 - 0.67 \approx 1.3\; \rm m$ away from the $20 \times 10^{-19}\; \rm C$ charge.

8 0

3 years ago

Calculate the electric field at the center of a square 46.4 cm on a side, if one corner is occupied by a +42.0 µc charge and the

liraira [26]

centre of square disrance to each corner found by Pythagoras' theorem.

coulombs law used to clculate field of each charge at centre

fields added vectorially for res

8 0

3 years ago