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3 years ago

Find the equivalent resistance of this parallel circuit with two strands.

1 answer:

6 0

In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then
1/Rt = 1/7 + 1/49
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 Ω

The total resistance hence is 49/8 Ω

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A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

Debora [2.8K]

<h3>Answer;</h3>

The period of the wave is 4 seconds

<h3>Explanation;</h3>

The period of a wave or periodic time is the time taken for one complete oscillation to occur. In this case, one complete oscillation occurs when the wave moves from one crest to the next or a trough to the next. This takes 4 seconds. Therefore the period is 4 seconds.
Frequency on the other hand is the number of oscillations by a wave in one second. Thus, f = 1/T, that is frequency is the reciprocal of periodic time.

4 0

3 years ago

Read 2 more answers

A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

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

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is: