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3 years ago

What direction does the coriolis effect spin in the southern hemisphere?

1 answer:

5 0

<span>"The direction of motion is caused by the Coriolis effect. This can be ... storms in the Northern Hemisphere, but rotate clockwise in the <span>Southern Hemisphere</span></span>

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WILL GIVE YOU BRAINLIST IF YOU ANSWER Which of the following characteristics of the Arctic rabbit is specifically an adaptation

andrew-mc [135]

Answer:

Hold active layer of soil in place; act as producers in ecosystem

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3 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

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

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3 years ago

An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in th

Harlamova29_29 [7]

Answer: $1.816\times 10^{-19} J$

Explanation:

Given

$E=\frac{hc}{\lambda }$

$E=2.18\times 10^{-18}(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

where h=Planck constant

c=speed of light

$E=2.18\times 10^{-18}(\frac{1}{3^2}-\frac{1}{6^2})$

$E=2.18\times 10^{-18}\times \frac{1}{12}$

$E=1.816\times 10^{-19} J$

7 0

3 years ago

Which force links electricity and magnetism together?

xenn [34]

Answer:

the electromagnetic force

4 0

3 years ago

An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of space between two parallel plates, starting a

Archy [21]

Answer:

Explanation:

Given that

speed u=4×10⁶ m/s

electric field E=4×10² N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates

acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s

now we find the horizantal distance travelled by electrons hit the plates

horizantal distance X=u[2y/a]^1/2

=4×10⁶[2×2×10⁻²/7×10¹³]^1/2

=9.5cm

now we find the velocity f the electron strike the plate

v²-(4×10⁶)²=2×7×10¹³×2×10⁻²

v²=16×10¹²+28×10¹¹

v²=1.88×10¹³m/s

speed after hits =>V=4.34×10⁶ m/s

7 0

3 years ago