Which process do hydrogen atoms use to make the Sun's energy?

A kite is 100m above the ground. if there are 200m of string out what is the angle between the string and the horizontal? (assum

jeka57 [31]

Answer : The angle between the string and the horizontal is 30 degrees

Explanation: Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .

Let the angle between the string and the horizontal be theta.

Now sin (Theta) = opposite side/hypotenuse

= 100/200 = 1/2

Therefore Theta = Sin ⁻¹ ( 1/2 )

Theta = 30 degrees

4 0

3 years ago

What is it that makes a magnet different from a piece of iron that is not mangetic?

Alisiya [41]

Answer:

Under normal conditions, a magnetic material like iron doesn't behave like a magnet because the domains don't have a preferred direction of alignment. On the other hand, the domains of a magnet (or a magnetized iron) are all aligned in s specific direction.

7 0

2 years ago

During a normal reaction to a stressful event, muscles are moved to their maximum capacity, and sensitivity is

Aleonysh [2.5K]

Answer:

The paper focuses on the biology of stress and resilience and their biomarkers in humans from the system science perspective. A stressor pushes the physiological system away from its baseline state toward a lower utility state. The physiological system may return toward the original state in one attractor basin but may be shifted to a state in another, lower utility attractor basin. While some physiological changes induced by stressors may benefit health, there is often a chronic wear and tear cost due to implementing changes to enable the return of the system to its baseline state and maintain itself in the high utility baseline attractor basin following repeated perturbations. This cost, also called allostatic load, is the utility reduction associated with both a change in state and with alterations in the attractor basin that affect system responses following future perturbations. This added cost can increase the time course of the return to baseline or the likelihood of moving into a different attractor basin following a perturbation. Opposite to this is the system's resilience which influences its ability to return to the high utility attractor basin following a perturbation by increasing the likelihood and/or speed of returning to the baseline state following a stressor. This review paper is a qualitative systematic review; it covers areas most relevant for moving the stress and resilience field forward from a more quantitative and neuroscientific perspective.

Explanation:

8 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

A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc

liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

v₀ₓ = 10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

v_y = $v_{oy}$ - g t

$v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)$

at the highest point of the trajectory the vertical speed must be zero

v_y = 0

therefore the velocity of the body is

v = (10 i ^ + 0j ^) m / s

the acceleration is

a = (0 i ^ - g j⁾

a = (0i ^ - 9.8 j ^) m / s²

5 0

2 years ago