Where is the majority of the mass located in an atom

Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential di

GenaCL600 [577]

(a) We can find the current flowing between the walls by using Ohm's law:
$I= \frac{\Delta V}{R}$
where $\Delta V=69 mV=0.069 V$ is the potential difference and $R=6.8\cdot 10^9 \Omega$ is the resistance. Substituting these values, we get
$I=1.01 \cdot 10^{-11} A$

(b) The total charge flowing between the walls is the product between the current and the time interval:
$Q=I \Delta t$
The problem says $\Delta t=0.86 s$ , so the total charge is
$Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C$

The current consists of Na+ ions, each of them having a charge of $e=1.6 \cdot 10^{-19} C$ . To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
$N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions$

4 0

2 years ago

You do 20 J of work pushing a crate up a ramp. If the output work from the inclined plane is 11 J, then what is the efficiency o

fomenos

Answer:

55%

Explanation:

take efficiency=power output/power input multiply by 100%

5 0

3 years ago

What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

sladkih [1.3K]

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0

2 years ago

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov

Brums [2.3K]

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

6 0

3 years ago

The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a

Setler [38]

Answer:

x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

$x_{cm}$ = 1 / $M_{total}$ ∑ $x_{i} m_{i}$

Where M_{total} is the total masses of the system, $x_{i}$ is the distance between the particles and $m_{i}$ is the masses of each body

Let's apply this equation to our problem

M = Me + m

M = 5.98 10²⁴ + 7.36 10²²

M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]

x_{cm} = 4.644 10⁶ m

4 0

3 years ago