Which imaging technique uses radiation that is emitted from the material being imaged?

If an amount of heat Q is needed to increase the temperature of a solid metal sphere of diameter D from 4°C to 7°C, the amount o

Hitman42 [59]

Answer:

Q = c M ΔT where c is the heat capacity and M the mass present

Q2 / Q1 = M2 / M1 since the other factors are the same

M = ρ V where ρ is the density

M = ρ Π (d / 2)^2 where d is the diameter of the sphere

M2 / M1 = (2 D/2)^2 / (D/2)^2 = 4

It will take 4Q heat to heat the second sphere

7 0

2 years ago

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

What is the effect of using split rings in a simple d.c generator

Aliun [14]

Answer:

Hence, The effect of using split rings in a simple DC motor is that the direction of the current flowing in the coil is reversed.

5 0

3 years ago

Explain why science is a continuous progression of study

Alenkinab [10]

new discoveries lead to new research.

Lise Meitner split the nucleus => nuclear energy research

8 0

3 years ago

Read 2 more answers

The following figure is from a student’s experiment to find the centre of mass of an irregular shape.

ANEK [815]

To find the center of mass of an irregular shape using

String- for hanging at the center

• A small mass - hung from the string

• A stand-holding up the string

• A pencil pinpointing the center

• A straight-edge- Identify the middle

<h3>What is the center of mass?</h3>

Generally, a point indicates the mean location of the matter in a body or system.

In conclusion, the center of mass also indicates a point at which the body is most balanced under gravity.