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GrogVix [38]
2 years ago
5

A pendulum of 50 cm long consists of small ball of 2kg starts swinging down from height of 45cm at rest. the ball swings down an

d strikes a bigger ball. what is the maximum kinetic energy of the 2kg bob
Physics
1 answer:
Ket [755]2 years ago
5 0

Assuming that all energy of the small ball is transferred to the bigger ball upon impact, then we can say that:

Potential Energy of the small ball = Kinetic Energy of the bigger ball

Potential Energy = mass * gravity * height

Since the small ball start at 45 cm, then the height covered during the swinging movement is only:

height = 50 cm – 45 cm = 5 cm = 0.05 m

Calculating for Potential Energy, PE:

PE = 2 kg * 9.8 m / s^2 * 0.05 m = 0.98 J

Therefore, maximum kinetic energy of the bigger ball is:

<span>Max KE = PE = 0.98 J</span>

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How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t
hodyreva [135]

Answer:

The heat transferred  from water to skin  is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred  from water to skin

Using formula for stream

Q=mc\Delta T

Put the value into the formula

Q=25\times10^{-3}\times4190\times(373-307)

Q=6913.5\ J

Hence, The heat transferred  from water to skin  is 6913.5 J.

3 0
2 years ago
Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12
Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-

In case all three capacitors are connected in parallel, the 2.0\;\mu\text{F} capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-

In case the 2.0\;\mu\text{F} capacitor is connected in parallel with the 1.5\;\mu\text{F} capacitor, and the two capacitors in parallel is connected to the 3.0\;\mu\text{F} capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}.

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}.

Voltage is the same across two capacitors in parallel.As a result,

\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}.

<h3>Case C: 2.76 V.</h3>

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Similarly,

  • the effective capacitance of the two capacitors in parallel is 5.0 μF;
  • the effective capacitance of the three capacitors, combined: \displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}.

Charge stored:

Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}.

Voltage:

\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}.

<h3 /><h3>Case D: 4.00 V</h3>

-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-.

Connect all three capacitors in series.

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}.

For each of the three capacitors:

Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}.

For the 2.0\;\mu\text{F} capacitor:

\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}.

6 0
2 years ago
If the top circuit has an oscillation frequency of 1000 Hz, the frequency of the bottom circuit is:_______.
kiruha [24]

Answer:

1410 Hz

Explanation:

Capacitance is reduced by 2, so the angular frequency will increase by a factor of \sqrt{2}.

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Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
DedPeter [7]
It will be <span>Einstein's General Relativity</span>
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trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

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