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2 years ago

Why can you hear a man's voice around the corner of a building even if you can't see him?

Physics

1 answer:

OverLord2011 [107]2 years ago

4 0

Answer:

Diffraction

Explanation:

the sound bounces off the corner and we are able to hear it

A P E X

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Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t

zalisa [80]

Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

5 0

2 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

3 years ago

Each second, 1250 m3 of water passes over a waterfall 150 m high. Three-fourths of the kinetic energy gained by the water in fal

mote1985 [20]

Answer:

The generator produces electrical energy at a rate of 1378125000 J per second.

Explanation:

volume of water falling each second is 1250 $m^{3}$

height through which it falls, h is 150 m

mass of 1 $m^{3}$ of water is 1000 kg

⇒mass of 1250 $m^{3}$ of water, m = 1250×1000 = 1250000 kg

acceleration due to gravity, g = 9.8 $\frac{m}{sec^{2} }$

in falling through 150 m in each second, by Work-Energy Theorem:

Kinetic Energy(KE) gained by it = Potential Energy(PE) lost by it

⇒KE = mgh

= 1250000×9.8×150 J

= 1837500000 J

Electrical Energy = $\frac{3}{4}$ (KE)

= $\frac{3}{4}$ ×1837500000

= <u>1378125000 J per second</u>

8 0

2 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

3 years ago

using hooke's law, f spring = k triangle x, find the elastic constant of a spring that stretches 2 cm when a 4 newton force is a

Ksivusya [100]

As we know that spring force is given as

$F = kx$

here we know that

F = 4 N

x = 2 cm = 0.02 m

now from the above equation we will have

$4 = k(0.02)$

$k = 200 N/m$

so the elastic constant of the spring will be 200 N/m

8 0

3 years ago