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swat32
2 years ago
14

What can you do to prevent an echo and a reverberation in cinema halls and auditorium?​

Physics
1 answer:
Gemiola [76]2 years ago
7 0

Answer:

Just make sure the distance between two objects isn't much

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Using two boxes, design two solar ovens. Before you build, describe your designs. Discuss elements such as materials, box shape,
Marizza181 [45]

Answer:

Find two boxes just a bit smaller than the other. The long sides of the box should be less than twice as long as the short side of the boxes. The smaller box should fit inside the larger box with about 1 inch in each direction to spare. The boxes can be cut down so that they fit together properly. Leave the flaps on the boxes. Buy a small sheet of Plexiglas (tm) a little bit smaller than the width and length of the top of the box. You will also need four pieces of cardboard to use for reflectors.

Explanation:

3 0
2 years ago
Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti
Triss [41]

Answer:

<em>The total potential (magnitude only) is 11045.45 V</em>

Explanation:

<u>Electric Potential </u>

The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by

\displaystyle V=\frac{kq}{r}

Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

V_2=-22090.91\ V

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

V_3=+22090.91\ V

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V

The total electric potential in A is

V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V

V=-11045.45 \ V

The total potential (magnitude only) is 11045.45 V

8 0
3 years ago
Rusting metal is an example of a _________ change. A) phase B) state C) chemical D) physical
sergeinik [125]

Answer:

C. Chemical Change

Explanation:

6 0
3 years ago
Read 2 more answers
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
3 years ago
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
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