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2 years ago

What are the products of a fusion reaction? Check all that apply.

Physics

1 answer:

Free_Kalibri [48]2 years ago

8 0

Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.

<h3>What are the products of a fusion reaction?</h3>

Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.

So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.

Learn more about reaction here: brainly.com/question/11231920

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Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

The diagram below shows a food web.

Mekhanik [1.2K]

i think it’s B. sorry if i’m wrong

8 0

3 years ago

Read 2 more answers

An object has a mass of 30 grams and measures 3cmx2cmx1cm. What is the density of the object?

dimulka [17.4K]

The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,

V = L x W x H

Substituting,

V = (3 cm)(2 cm)(1 cm) = 6 cm³

From the idea presented above,
d = m/V

Substituting the known values,

d = (30 g)/ (6 cm³) = 5 g/cm³

ANSWER: 5 g/cm³

4 0

3 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

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

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

5 0

3 years ago