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1 year ago

12

A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period o

f the motion

1 answer:

Jobisdone [24]1 year ago

7 0

Answer:

Explanation:

The amplitude is extraneous information for this question.

Period (T) is simply a measure of time (how long it takes to complete a full cycle).

The frequency (f), is the reciprocal (how many cycles it can complete per unit of time).

The frequency reported is 12Hz. Hz is a unit meaning "cycles per second"

Thus 12cycles = 1second.

To find the number of seconds per cycle, simply create the other unit fraction and simplify:

$T=\frac{1second}{12cycles}\\T=0.8\bar{3}\frac{seconds}{cycle}\\$

$T=0.83seconds$

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A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

2 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

$B'=\dfrac{\mu_{0}I}{2\pi r}$

Where, r = distance

I = current

Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

$B'=1.8\times10^{-5}\ T$

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

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3 years ago

What depends on public opinion and support?

Lelechka [254]

Answer: A.) Regulations at a local level.

6 0

2 years ago

you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t

telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

velocity of the ant, v = 2 m/s

change in position of the ant, Δx = 0.81 m

initial time, t₁ = 4 s

time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

$v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s$

The time elapsed since you stopped the stopwatch is calculated as;

$t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s$

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

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3 years ago

Qué representa el lanzamiento vertical ascendente

rusak2 [61]

Answer:

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2 years ago