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3 years ago

A red blood cell contains 4.8 107 free electrons. What is the total charge of these electrons in the red blood cell?

Physics

1 answer:

tatuchka [14]3 years ago

7 0

Answer:

Charge, $q=7.68\times 10^{-12}\ C$

Explanation:

It is given that,

The number of electron in a RBCs, $n=4.8\times 10^7$

We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :

q = ne

e is the change on electron

$q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C$

So, the net charge is $7.68\times 10^{-12}\ C$ .

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The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

A sound wave traveling eastward through air

Wewaii [24]

A sound wave traveling eastward through air causes the air molecules to Vibrate east and west.

5 0

3 years ago

Read 2 more answers

A body is dropped from the roof of a 20 m high building by how much:

USPshnik [31]

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

Height (h) = 20 m
Gravity (g) = 9.8 m/s²
Time (t) = ?
Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

$\boxed{t=\sqrt{\frac{2*h}{g}}}$

Replace:

$\boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}$

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

$\boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}$

It divides:

$\boxed{t=\sqrt{4.08s}}$

The square root is performed:

$\boxed{t=2.01s}$

==================================================================

Final Velocity

use formula:

Vf = g * t

Replace:

Vf = 9.8 m/s² * 2.01 s

Multiply:

Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in 2.01 seconds.

How fast does it hit the ground?

Hits the ground with a speed of 19.7 meters per seconds.

7 0

3 years ago

9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

$W = E_p = mgh$

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

$W = 50*9.8*13 = 6370 J$

3 0

3 years ago

What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?

Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ .

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

$p=mv$

$p=3400\ kg\times 201.16 \ m/s$

$p=683944\ kg-m/s$

$p=6.83\times 10^5\ kg-m/s$

So, the magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ . Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers