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drek231 [11]
3 years ago
6

An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible

internal resistance. What is the initial current in the circuit, expressed in milliamperes? Calculate the circuit\'s time constant in milliseconds. How much time, in milliseconds, must elapse from the closing of the circuit for the current to decrease to 2.57% of its initial value?
Physics
1 answer:
harkovskaia [24]3 years ago
3 0

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

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1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
We cannot consider something to be work unless
iren [92.7K]
<span>force applied causes movement of an object in the same direction as the applied force.</span>
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Read 2 more answers
A materials density is the same , no matter how large or small the sample is or what it’s shape is as long as it is solid unifor
Scorpion4ik [409]

Answer:

See the explanation below

Explanation:

Density is defined as the relationship between mass and volume, i.e. the following equation can be used:

density = m/v

where:

density [kg/m^3]

m = mass [kg]

v = volume [m^3]

If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.

m = density*v

To understand this concept more clearly, let's use the following example:

We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.

1000 = m /1

m = 1000*1 = 1000 [kg]

Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.

1000 = 500/v

v = 500/1000

v = 0.5 [m^3]

We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.

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3 years ago
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Car with a mass of 1210 kg moving at a velocity of 51 m/s.
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Hope i helped
Have a good day :)

 
6 0
3 years ago
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