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2 years ago

What unit is used to measure the period of a wave?

Physics

1 answer:

mixer [17]2 years ago

4 0

Answer:

D. Meters/Seconds

Explanation:

The time period of a wave is measured in seconds.

A typical wave involves both time and distance. Consider a sound wave, which is basically a periodic modulation of the local air pressure. We "hear" the sound because our ears respond to the variations of pressure.

The most common metric of a sound wave is frequency. This is the rate at which the change in pressure occurs, and is measured in cycles per second, formally known as "hertz". The period is the inverse of frequency andl has the units of seconds per cycle, commonly stated simply as seconds.

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A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows th

ddd [48]

Answer:

$8\mu C$

Explanation:

t = Time taken = $2\mu s$

i = Current = 3 A

q(0) = Initial charge = $2\mu C$

Charge is given by

$q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C$

The magnitude of the net electric charge of the capacitor is $8\mu C$

4 0

3 years ago

A sample of radium-226 will decay to ¼ of its original amount after 3200 years. What is the half-life of radium-226?

lesantik [10]

<h3><u>Answer</u>;</h3>

1600 years

<h3><u>Explanation</u>;</h3>

Half life is the time taken for a radioactive isotope to decay by half of its original amount.
We can use the formula; N = O × (1/2)^n ; where N is the new mass, O is the original amount and n is the number of half lives.
A sample of radium-226 takes 3200 years to decay to 1/4 of its original amount.

Therefore;

Thus; <em>3200 years is equivalent to 2 half lives.</em>

<em>Hence, the half life of radium-226 is 1600 years</em>

3 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t

NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

5 0

3 years ago

Two skaters stand facing each other. One skater's mass is 60 kg, and the other's

jeyben [28]

Answer:

v' = 2.4 m/s

Explanation:

Given that,

Mass of one skater, m = 60 kg

Mass of the other's skater, m' = 60 kg

The two skaters push off each other. After the push, the smaller skater has a velocity of 3.0 m/s.

When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.

mv = m'v'

$v'=\dfrac{mv}{m'}\\\\v'=\dfrac{60\times 3}{75}\\\\v'=2.4\ m/s$

So, the velocity of the larger skater is 2.4 m/s.

3 0

2 years ago