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2 years ago

7th grade science i mark as brainliest

Physics

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

YOU ARE NOOB BECAUSE YOU ARE FREE FIRE PLAYER

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if you have several different substances mixed together, like Italian salad dressing, where do the most dense materials go where

Alecsey [184]

If you've ever mixed water and oil together,you've probably noticed that the oil rises to the top and the water stays at the bottom-- no matter how much you try to shake them. This is because oil is less dense than water, making it float to the top. The molecules in oil are larger than those in water, so they don't pack as tightly together as they do in water. In a mixture of oil and vinegar, the<span> </span>oil <span>floats on the </span>vinegar/<span>water mixture, while the solids sink to the bottom. So, if one material floats over another, the material on top is less dense.

Let me know if this helps! :)</span>

5 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

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car

gogolik [260]

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

V_i = 110m/s
V_f = 80m/s
t_i = 0s
t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

4 0

3 years ago

Can a body having larger mass be hotter than another of smaller mass.expalain

Rom4ik [11]

Answer:

it would make sense because a larger body could produce more body heat.

8 0

3 years ago