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IRISSAK [1]
3 years ago
11

A 15-µF capacitor is charged to 40 V and then connected across an initially uncharged 25-µF capacitor. What is the final potenti

al difference across the 25-µF capacitor?
Physics
2 answers:
allochka39001 [22]3 years ago
8 0

Answer:

40V

Explanation:

The potential difference across the initially uncharged capacitor is still 40V since there both connected in parallel. For parallel capacitor connections, the potential difference across the terminals (ends) of the capacitors is the same for all the capacitors but the charge on each capacitor depends on it capacitance.

So in this case the uncharged capacitor is in a parallel connection with the charged 40V capacitor. The connection of 15μF capacitor across the uncharged capacitor sets up a potential difference of 40V across its ends.

Kobotan [32]3 years ago
6 0

Answer:

24volts

Explanation:

If a 15-µF capacitor is charged to 40V, the charge across the capacitor can be calculated using the formula;

Q = CV where;

Q is the charge flowing across the capacitor

C is the capacitance of the capacitor. = 15-µF

V is the voltage = 40V

Q = 15×10^-6×40

Q = 0.0006coulombs

If the charge of 0.0006coulombs is then connected across an initially uncharged 25-µF capacitor, the potential difference across the 25-µF can be calculated using the initial expression;

Q = CV

V = Q/C

V = 0.0006/25×10^-6

V = 24Volts

Therefore, the final potential difference across the 25-µF capacitor will be 24volts

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A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
3 years ago
A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a
anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = 4.83 \AA

Intercepts along the three axes:

\bar{x} = 9.66 \AA

\bar{x} = 19.32 \AA

\bar{x} = 14.49 \AA

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = \frac{1}{9.66}

b = \frac{1}{19.32}

c = \frac{1}{14.49}

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = \frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}

b' = \frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}

c' = \frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}

7 0
3 years ago
Anybody have the answer?
Anika [276]
IT IS C
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3 0
4 years ago
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Which is not part of the first law of thermodynamics?
Nataly [62]
C. Energy can change forms
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4 years ago
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The material is called paramagnetic or non magnetic material. Opposite to this are materials whose magnetic domains are aligned and possess asmagnetism which is due to its ferrous or iron content.
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