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3 years ago

Sometimes it is very hard to hear in a gymnasium because of echoes . How can you changed it

1 answer:

7 0

A couple ways to reduce the bouncing of sound waves are to put foam in the wall and put slits down the wall so the sound is asorbed, another way is to hang soft mats of foam from the ceiling, that also asorbs the rogue sound waves.

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Why did the following change occur? the snow that covered the ground all winter began to melt, temperatures rose, and flowers st

FromTheMoon [43]

Answer:C

Explanation:

8 0

2 years ago

Read 2 more answers

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

3 years ago

A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e Newton's first law of motion.

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

8 0

3 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0

3 years ago

When there is no number in front of a chemical formula in a chemical equation, what number is understood?

Andre45 [30]

The number in front is the number of molecules (or atoms) taking part in the (balanced) chemical reaction equation.

5 0

3 years ago