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4 years ago

How are changes in energy and matter related

Physics

1 answer:

Svetlanka [38]4 years ago

5 0

Answer:

According to Einstein's famous equation, matter can convert into energy (and viceversa) as follows:

$E=mc^2$

where

E is the energy

m is the mass

c is the speed of light ( $3\cdot 10^8 m/s$ )

Given the huge value of $c^2$ , we see that even a tiny amount of matter is able to release a huge amount of energy, when the whole mass is converted into energy. This is precisely what happens in nuclear reactions. For example, in the process of nuclear fusion (that occurs in the core of the stars), two light nuclei fuse together into a heavier nucleus. The mass of the final nucleus is lower than the total mass of the initial nuclei, so part of the mass has been converted into energy according to the equation above: this is why the amount of energy produced by stars is so big.

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The two parents shown in (Figure 1) exert upward forces

timama [110]

Answer:

W = 83 [N] (weight)

Explanation:

We must clarify that the system is in static equilibrium, in such a way that we must perform a sum of forces on the Y-axis equal to zero, in order to determine the downward force corresponding to the weight of the child.

∑Fy = 0

$F_{1}+F_{2}-m*g=0\\18+65=m*g\\m*g = 83 [N]$

And the mass can be determined as follows:

$m*g=83\\m = 83/9.81\\m=8.46[kg]$

5 0

3 years ago

Why cant your rocket could never reach the speed of light?

Oksanka [162]

Answer:

The length of the object would shrink to zero which is not possible.

Explanation:

A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.

The Lorentz factor is given as:

$\gamma=\frac{1}{\sqrt{\frac{v^2}{c^2} } }$

where:

v = speed of the moving object

c = speed of light

4 0

3 years ago

A proton is located at <3 x 10^-10> m. What is r, the vector from the origin to the location of the proton

Eduardwww [97]

Complete Question

A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton

Answer:

The vector position is $\=r=$

Explanation:

From the question we are told that

The position of the proton is $m$

Generally the vector location of the proton is mathematically represented as

$\= r =$

So substituting values

$\=r=$

4 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago

You are on a cruise ship traveling north at a speed of 13 m/s with respect to land. 1) if you walk north toward the front of the

Brrunno [24]

In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.

Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.

The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.

In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.

Relative velocity = Velocity of Body A+ Velocity of Body B.

Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)

Our own velocity headed northwards(B)= 2.8 m/s.

Relative velocity = Velocity of Body A+ Velocity of Body B.

Relative velocity= 13.0 + 2.8 = 15.8m/s.

Thus our own velocity with respect to the land is 15.8 m/s.

5 0

3 years ago