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3 years ago

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A raft is made of 14 logs lashed together. Each log is 42 cm in diameter and a length of 6.4 m. 42% of the log volume is above t

he water when no one is on the raft. Determine the following: the specific gravity of the logs.

1 answer:

Pavel [41]3 years ago

7 0

Answer:

Explanation:

Given

No of logs $n=14$

diameter of log $d=42\ cm$

Length of log $L=6.4\ m$

42 % of log volume(V) is above water when no one is on raft

so 58 % of log volume(V) is submerged in the water

Weight of 14 log

$W=14\times \rho _{log}\times V\times g$

Buoyancy force on 14 logs $F_b=14\times \rho _{water}\times 0.58V\times g$

as system is in equilibrium so

$W=F_b$

$14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g$

$\rho _{log}=0.58\rho _{water}$

$\frac{\rho _{log}}{\rho _{water}}=0.58$

Specific gravity of log $=0.58$

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The rate at which speed changes is called

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Answer:

meters per second

Explanation:

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3 years ago

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While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega

Paraphin [41]

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

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3 years ago

A Man Moved first a Distance of 1000 m in 25 second and 2.5 km in 50 second along a in straight line?

11Alexandr11 [23.1K]

Answer:

Average speed = 46.67 m/s

Explanation:

Given that the time taken in covering first 1000 m = 25 seconds.

The time taken in covering next 2.5 km = 50 seconds.

Total distance covered = 1000 m + 2500 m = 3500 m

Total time taken = 25+50=75 seconds

Average speed = Total distance covered / total time taken

= 3500/75 = 46.67 m/s

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3 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

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3 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

3 years ago