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3 years ago

10

Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4

kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based on the data given below. Total energy is the sum of gravitational potential energy and kinetic energy. In this problem, round gravity to: g = 10 m/s^2.

1 answer:

Lerok [7]3 years ago

3 0

Answer:

its should be 2.0 and 4.5 on it

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A. Reduced greenhouse gas emissions.

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Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin

Flura [38]

Answer:

<em>a. 0.33 Amp</em>

<em>b. 2.4 Volt</em>

<em>c. 0</em>

<em>d. 2.45 Amp</em>

<em>e. Infinite</em>

<em>f. Series is safer</em>

Explanation:

<u>Series Connection of Resistors</u>

When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.

a. The string of 50 bulbs is connected to a 120 VAC outlet and consumes 40 W. The power of a circuit is given by

$P=V.I$

Solving for I

$\displaystyle I=\frac{P}{V}=\frac{40}{120}=0.33\ Amp$

Since all the bulbs are connected in series the current is the same for all of them.

b. The voltage is equally divided, so each bulb has 120/50= 2.4 V

c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.

d. If one of the bulbs short circuits, the resistance of that bulb is zero and the voltage is distributed by the 49 remaining bulbs. Thus the new current is

$\displaystyle I=\frac{V}{R}=\frac{120}{49}=2.45\ A$

e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.

f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.

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3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

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