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pychu [463]
3 years ago
11

Which would not provide a useful measurement of temperature?

Physics
1 answer:
STatiana [176]3 years ago
6 0

It could maybe be D which is the expansion of a liquid inside a closed tube

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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
A mover applies a net force of 28 N to a sofa that has a mass of 70 kg.
Alinara [238K]

Answer: .4 m/s^2= acceleration

Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

8 0
3 years ago
Can I have help with this please- thank you!!!
seropon [69]

Answer:

I would say there is friction against the floor, air resistance, and gravity.

Explanation:

6 0
3 years ago
A very long straight current-carrying wire produces a magnetic field of 20 mt at a distance d from the wire. to measure a field
padilas [110]

The distance covered is 4d.

<h3>What is the distance?</h3>
  • Distance is a measure of how far apart two objects or locations are using numbers.
  • The distance can refer to a physical length or an estimate based on other factors in both daily language and physics (e.g. "two counties over").
  • The terms "distance from A to B" and "distance from B to A" are frequently used interchangeably.
  • A distance function or metric is a technique to describe what it means for components of space to be "near to" or "far away" from each other in mathematics.
  • It is a generalization of the idea of physical distance. Distance is a non-numerical unit of measurement in psychology and the social sciences; psychological distance is the separation of an object from itself along dimensions including time, space, and isolation.

We know that

the magnetic field is inversely proportional to distance.

\dfrac{B_1}{B_2} = \dfrac{D_2}{D_1}

\dfrac{20}{5} = \dfrac{D_2}{d}

D_2 = 4d

To learn more about distance with the given link

brainly.com/question/15172156

#SPJ4

3 0
2 years ago
What Device can be used to increase voltage from a source of direct-current?
Rudiy27
Multi meter reading instruments Device can be used to increase voltage know direct-current
4 0
3 years ago
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