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1 year ago

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A simple pendulum of length 1.5m has a bob of mass 2.0kg. State the formula for the period of small oscillations and evaluate it

in this case. (2.43s)

1 answer:

Shalnov [3]1 year ago

4 0

A simple pendulum of length 1.5m has a bob of mass of 2.0kg, then the period for the small oscillations would be 2.456 s.

<h3>What is the frequency?</h3>

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,

f = 1/T

As given in the problem A simple pendulum of length 1.5m has a bob of mass of 2.0kg.

The formula for the time period for the pendulum,

T = 2π√L/g

=2π√1.5/9.81

=2.456 s

Thus, the period for the small oscillations would be 2.456 s.

Learn more about frequency from here,

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Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an un

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Answer: $316.8V$

Explanation:

given data:

metre moving current = $0.96mA$

meters voltage = $288V$

or $0.96*300V = 288V$

<u><em>Solution:</em></u>

<u><em /></u> $v1 = (0.96mA*150)$ <u><em /></u>

<u><em /></u> $= 144V$ <u><em /></u>

<u><em /></u>

$i1 = \frac{144v}{750}$

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$i2 = imovement + i1$

$i2= 0.96mA+0.192mA$

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3 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

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In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.

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Answer:

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When the size increase, the distance between the nuclei also increase leading a decrease the force of attraction between the nuclei

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