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3 years ago

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What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 c

m from the other) is 293 V?

1 answer:

posledela3 years ago

6 0

Answer:

E = 4156.02 Vm⁻¹

Explanation:

The magnitude of the uniform electric field between the plates can be given by the following formula:

$E = \frac{\Delta V}{d}\\$

where,

E = Electric field strength = ?

ΔV = Potetial Difference = 293 V

d = distance between plates = 7.05 cm = 0.0705 m

Therefore,

$E = \frac{293\ V}{0.0705\ m}\\\\$

<u>E = 4156.02 Vm⁻¹</u>

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Elan Coil [88]

• The only force acting on the ball, and thus the net force, is due to its own weight, with magnitude

<em>w</em> = <em>m</em> <em>g</em> = (0.4 kg) (9.8 m/s²) ≈ 3.9 N

pointing downward.

• The ball is in free fall, so its acceleration is <em>g</em> = 9.8 m/s² (also pointing downward).

• The ball is dropped from a height of 49 m, so its height <em>y</em> at time <em>t</em> is

<em>y</em> = 49 m - 1/2 <em>g</em> <em>t</em> ²

Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 49 m - 1/2 <em>g t</em> ²

<em>t</em> ² = (98 m) / <em>g</em>

<em>t</em> = √((98 m) / <em>g</em>) = √(10) s ≈ 3.2 s

• The ball's velocity <em>v</em> at time <em>t</em> is

<em>v</em> = - <em>g t</em>

so that at the time found previously, the ball will have attained a velocity of

<em>v</em> = - <em>g</em> (3.2 s) ≈ -31 m/s

and thus a <em>speed</em> of about 31 m/s.

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3 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

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3 years ago

of the waves listed, the one with the longest wavelength is a(n) a. x ray. b. radio wave. c. microwave. d. visible light wave.

djyliett [7]

From the items on that particular list, the longest waves are radio waves.

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3 years ago

The freezing point of a solution prepared by dissolving 150 × 10–3 g of caffeine in 10.0 g of camphor is lower by 3.07°C than th

Finger [1]

Answer:

Molar Mass=193.9738 g/mol

Molar Mass≅194g/mol

Explanation:

Consider the formula:

m=ΔT/K_{f}

where:

ΔT is freezing point depression

K_{f} is freezing point depression constant

m is the morality=(moles of solute/kg of solvent)

Now:

$m=\frac{3.07}{39.7}$

$m=0.07733 moles of solute/kg of solvent$

Now:

0.07733 (moles of solute/kg of solvent) *0.010 kg of solvent

$7.733*10^{-4}$ moles of solute(Caffeine)

Molar mass = Mass of solute/moles of solute

Molar Mass= $\frac{150*10^-3}{7.733*10^{-4} }$

Molar Mass=193.9738 g/mol

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3 years ago

Aluminum rivets used in airplane construction are made slightly larger than the rivet holes and cooled by "dry ice" (solid CO2)

poizon [28]

Answer:

$\rm 4.554\ m m.$

Explanation:

<u>Given:</u>

Initial temperature, $\rm T_i=23.0\ ^\circ C=23 + 273.15 = 296.15\ K.$
Final temperature, $\rm T_f = -78.0\ ^\circ C = -78+273.15=195.15\ K.$
Diameter of the hole, $\rm d = 4.500\ m = 4.500\times 10^{-3}\ m.$
Expansion coefficient of the Aluminum, $\rm \alpha = 2.4\times 10^{-5}\ K^{-1}.$

Let the diameter of the rivet at $\rm T_i=23.0\ ^\circ C$ be $\rm d_o$ , such that,

$\rm d_o=d+\Delta d$

$\rm \Delta d$ is the elongation in the diameter of the rivet.

We know, The change in the diameter of the rivet is related with the change in temperature as:

$\rm \dfrac{\Delta d}{d_o}=\alpha \Delta T.$

where, $\rm \Delta T$ is the change in temperature = $\rm T_f-T_i=-195.15-296.15=-491.30\ K.$

Also, $\rm \Delta d=d_o-d.$

Using these values,

$\rm \dfrac{d_o-d}{d_o}=\alpha \Delta T\\1-\dfrac{d}{d_o}=\alpha \Delta T\\\dfrac{d}{d_o}=1+\alpha \Delta T\\\Rightarrow d_o = \dfrac{d}{1+\alpha \Delta T}\\=\dfrac{4.500\times 10^{-3}}{1+2.4\times10^{-5}\times (-491.30)}\\=4.554\times 10^{-3}\ m.\\=4.554\ m m.$

It is the required diameter of the rivet at $-78.0\ ^\circ C$ .

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3 years ago