If a weather system is moving from the Pacific coast to North Carolina,

A proton has been accelerated from rest through a potential difference of -1000 v . part a what is the proton's kinetic energy,

wlad13 [49]

We can apply the law of conservation of energy here. The total energy of the proton must remain constant, so the sum of the variation of electric potential energy and of kinetic energy of the proton must be zero:
$\Delta U + \Delta K=0$
which means
$\Delta K = - \Delta U$
The variation of electric potential energy is equal to the product between the charge of the proton (q=1eV) and the potential difference ( $\Delta V=-1000 V$ ):
$\Delta U = q \Delta V=(1 eV)(-1000 V)=-1000 eV$
Therefore, the kinetic energy gained by the proton is
$\Delta K = -(-1000 eV)=1000 eV$
<span>And since the initial kinetic energy of the proton was zero (it started from rest), then this 1000 eV corresponds to the final kinetic energy of the proton.</span>

4 0

3 years ago

Read 2 more answers

Two cars are moving in the same direction at the same speed of (72.0 kmh). What is

Licemer1 [7]

Answer:

0 km/h

Explanation:

Relative speed is the speed of a moving body with respect to another.

When two bodies move in the same direction then the relative speed is calculated as difference of their speeds.

In this case;

The two cars have the same speed. The relative speed will be;

72 km/h -72 km/h = 0 km/h

6 0

2 years ago

__________ is when all light rays bounce off an object at the same angle.

Ad libitum [116K]

Diffuse reflection have a great day

5 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

Which material reflects the most light??

gogolik [260]

The aluminum foil would reflect the most light !

7 0

3 years ago