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Alchen [17]
3 years ago
13

Pls help i will literally venmo you im not even kidding

Physics
1 answer:
Marizza181 [45]3 years ago
6 0

For Question 6:

The formula for momentum to express the conservation of momentum of two objects in a collision is:

m1 • v1 + m2 • v2 = m1 • v1' + m2 • v2'

m1 and m2 are the masses so

63v1 + 55v2 = 63v1' + 55v2'

v1 and v2 represent the velocities before the collision, so

(63)(0) + (55)(0) = 63v1' + 55v2'

v1' and v2' represent the velocities after the collision, so

(63)(0) + (55)(0) = (63)(3.3) + (55)v2' <-- We're looking for v2'

0 + 0 = 207.9 + 55v2'

-207.9 = 55v2'

-3.78 m/s = v2'

The answer is negative because the 55kg skater is moving to the left.

For Question 7:

The formula for momentum is p = m • v

so we have:

3.9 = .15v

26 m/s = v

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Answer: Amplitude

Explanation:

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3 years ago
A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci
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Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0
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A positively charged glass rod is bought close to a suspended metal needle. What
Darina [25.2K]

Answer:

attracted

Explanation:

opposite charges attract each other when the rub against each other

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From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
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An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
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Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

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With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

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                      U = 1.16 10⁻¹⁴ 1160 cos 0

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Antiparallel orientation

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                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

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3 years ago
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