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3 years ago

6

Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnect

ed and the capacitors are connected to a 33 ohm resistor, how long should it take for the voltage to cross the capacitors to drop to 2.2 volts (36.8% of the original 6.0 volts)?

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

$V = V_{o} e^{\frac{-t}{T}}$

$2.2 = 6 e^{\frac{-t}{66}}$

t = 66.2 sec

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Sladkaya [172]

Answer:

<em>The new pressure = 5.64 atm.</em>

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

<em>Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters</em>

<em>Substituting these values into equation 2,</em>

<em>P₂ = (2.5×900×336)/(298×450)</em>

P₂ = 756000/134100

P₂ = 5.64 atm.

<em>Thus the new pressure = 5.64 atm.</em>

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3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

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6 0

3 years ago

Joey is riding in an elevator which is accelerating upwards at 2.0 m/s2. The elevator weighs 300.0 kg, and Joey weighs 60.0 kg.

vodka [1.7K]

4200 N is the tension in the cable that pulls the elevator upwards.

The correct option is A.

<h3>What does tension ?</h3>

Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.

<h3>Briefing:</h3>

We employ the following formula to determine the cable's tension.

Formula:

T = mg+ma............ Equation 1

Where:

T is the cable's tension.

M = Mass of the elevator and the Joey

Accelerating with a

g = Gravitational acceleration

Considering the query,

Given:

m = (300+60) = 360 kg

a = 2 m/s²

g = 9.8 m/s²

Substitute these values into equation 2

T = (360×9.8)+(360×2)

T = 3528+720

T = 4248 N

T ≈ 4200 to the nearest hundred.

To know more about Tension visit:

brainly.com/question/14177858

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1 year ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

Orlov [11]

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

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$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

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4 years ago

According to the second law of thermodynamics, it is impossible for ____________. According to the second law of thermodynamics,

Tomtit [17]

Answer:

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Explanation:

Option A is incomplete and so it's possible.

Option B is possible

Option D is related to the first lae and has nothing to do with the second law.

Hence, the correct option is C.

The ideal engine follows a reversible cycle albeit an infinitely slow one. If the work is being done at this infinitely slow rate, the power of such an engine is zero.

We can also stat the second law of thermodynamics in this manner;

It is impossible to construct a cyclical heat engine whose sole effect is the continuous transfer of heat energy from a colder object to a hotter one.

This statement is known as second form or Clausius statement of the second law.

Thus, it is possible to construct a machine in which a heat flow from a colder to a hotter object is accompanied by another process, such as work input.

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3 years ago