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2 years ago

Which best explains how Selena can correct her error?

Physics

2 answers:

alekssr [168]2 years ago

8 0

Answer:

B:She can change the arrows so they show current traveling in opposite directions on the sides of the loop.

Explanation:

Scilla [17]2 years ago

7 0

Answer: well first you have to show or tell us the “error“ Selena made

Explanation:

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Which of the following can be used to induce abortion non surgically?

SSSSS [86.1K]

The answer is C.Mifepristone

7 0

2 years ago

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A bus travels from el Paso, texas, to chihuahua, mexico, in 5 hr with an average velocity if 75km/hr to the south. How far is Ch

katen-ka-za [31]

375km

Explanation:

Given parameters:

Time of travel = 5hr

Average velocity = 75km/hr south

Unknown:

Displacement between Chihuahua and El paso = ?

Solution:

Velocity is the rate of change of displacement of a body with time. It is a vector quantity that has both magnitude and direction;

Velocity = $\frac{displacement}{time}$

Since the unknown is displacement;

displacement = velocity x time

input the values;

displacement = 75km/hr x 5hr = 375km

learn more:

Velocity brainly.com/question/10883914

#learnwithBrainly

4 0

2 years ago

A 1-hectare field contains 2500 bean plants having a total leaf surface area of 557 m2. ambient temperature is 21Â°c, winds are

Mashcka [7]

Answer:

The field losses a total of 55.814 litres of water in an hour.

Explanation:

As the complete question is not given here, the average rate of water loss is not given here which is found in normal conditions as 1.67 mL/m2.

Now the average rate of water loss per minute is 1.67 mL/m2

So the average rate of water loss per minute for 557 m2 is 1.67*557=930.19 mL

So 930.19 mL of water is lost from the field in 1 minute, for the calculation of water lost in an hour, the value is given as 60*930.19 =55811.4 mL or 55.814 litres.

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3 years ago

Why does the layer of glass in a solar cell need an antireflective coating?

Lera25 [3.4K]

Answer:

The Anti Reflective Coating on a solar cells helps to increase the amount of light absorbed into the cell. This anti reflective coating is very much needed as the reflection of a bare silicon solar cells is over 30%. For the thin AR Coating, silicon nitride or titanium oxide is used.

Explanation:

hope this help:)

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2 years ago

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