Which of the following is an ecosystem

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a

DENIUS [597]

Answer:

C = 4,174 10³ V / m^{3/4} , E = 7.19 10² / ∛x, E = 1.5 10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

V = C $x^{4/3}$

C = V / x^{4/3}

C = 220 / (11 10⁻²)^{4/3}

C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

V = E dx

E = dx / V

E = ∫ dx / C x^{4/3}

E = 1 / C x^{-1/3} / (- 1/3)

E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

E = 3 / C (0- (-1 / x^{1/3}))

E = 3 / 4,174 10³ (1 / x^{1/3})

E = 7.19 10² / ∛x

for x = 0.110 cm

E = 7.19 10² /∛0.11

E = 1.5 10³ N/C

6 0

3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

or

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago

Which of the following represent units of capacitance? You may choose more than one correct answer:

inessss [21]

Answer:

Capacitance is a derived physical quantity measured in farad

4 0

2 years ago

A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J

REY [17]

1 nanowatt = 1 nanojoule/sec
1 watt = 1 joule/sec
10 watts = 10 joules/sec
100 watts = 100 joules/sec
742.914 watts = 742.914 joules/sec
1,000 watts = 1,000 joules/sec
10,000 watts = 10,000 joules/sec
100,000 watts = 100,000 joules/sec
1 megawatt = 1 megajoule/sec
1 gigawatt = 1 gigajoule/sec
1 petawatt = 1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0

3 years ago