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4 years ago

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An circuit has a section AB as shown in the fig. with E= 10V. (sf C_1) = 1.0 F, (sf C_2) = 2.0 F and the potential difference (s

f V_a-V_b = 5V). The voltage across (sf C_1) is?

1 answer:

bagirrra123 [75]4 years ago

4 0

Without the fig, we know nothing about AB, C_1, C_2, V_a, or V_b.
The probability of getting any kind of answer from anybody who hasn't
seen or tasted the fig is vanishingly small.

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

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A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

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2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

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