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2 years ago

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An electron is located in an electric field of magnitude 600. newtons per coulomb. what is the magnitude of the electrostatic fo

rce acting on the electron?

1 answer:

SOVA2 [1]2 years ago

6 0

The magnitude of the electrostatic force acting on a charge q is the product between the charge and the intensity of the electric field E. The magnitude of the electron charge is $q=1.6 \cdot 10^{-19}C$ (we are not interested in the sign), so the electrostatic force magnitude is
$F=qE=(1.6 \cdot 10^{-19}C)(600 N/C)=9.6\cdot 10^{-17}N$

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calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

1 year ago

Read 2 more answers

A cylinder of radius R and height H is floating upright in

emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

$F = mg$

so we have

$\Delta P \pi R^2 = mg$

so we have

$\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g$

so we have

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

6 0

2 years ago

during convection, hot material _____ (expands & sinks, cools & rises, expands & rises, deflates & rises) then m

Lady bird [3.3K]

Answer:

During convection, hot material expands & rises then moves to the side and cools & sinks. this circular pattern is called a convection current.

Explanation:

Convection is one of the three methods of transfer of heat. It occurs only in fluids (liquids or gases).

Convection occurs when there is a source of heat that heats a fluid, such as in a boiling pot of water. The water which is on the bottom of the pot becomes warmer before than the water at the top (because it is closer to the flame), and so it becomes less dense: for this reason, it expands and it becomes rising. On the contrary, the water on top is colder, so it is more dense and starts sinking, replacing the warmer water. As the new part of water gets warmer, it starts rising, and so the process is continuously repeated. This circular current is called convection current.

4 0

2 years ago

Read 2 more answers

There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the directi

wlad13 [49]

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (<em>in this case the -z direction</em>) and the curling of the fingers point in the direction of the magnetic field B→ (<em>in this case the +x direction</em>), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

6 0

1 year ago

Suppose that you measure the intensity of radiation from carbon-14 in an ancient piece of wood to be 6% of what it would be in a

borishaifa [10]

Answer:

t = 23136 years old

Explanation:

The intensity of radiation obeys the exponential decay law.

$I(t)=I_{0}e^{-\lambda t}$ (1)

Here:

λ is the radiation decay constant.

I is the intensity of radiation after t time.

I₀ is the initial intensity of radiation.

t is the time

Now, we know that the intensity of radiation is a 6% of what it would be in a freshly cut piece of wood, in other words:

$I(t)=0.06I_{0}$ (2)

Combining (2) with (1), we have:

$0.06I_{0}=I_{0}e^{-\lambda t}$

$0.06=e^{-\lambda t}$ (3)

Now, we just need to solve (3) for t.

$t=\frac{-ln(0.06)}{\lambda}$ (4)

and $\lambda = \frac{ln(2)}{t_{1/2}}$ (5)

$t_{1/2}$ is the half-life.

In our case, we have ¹⁴C, so the $t_{1/2} = 5700 y$

Finally, we can find t putting (5) in (4):

$t=\frac{-t_{1/2}ln(0.06)}{ln(2)} = 23136 y$

I hope it helps you!

6 0

2 years ago