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2 years ago

32. Increasing the amplitude of a sound wave produces a

Physics

1 answer:

ss7ja [257]2 years ago

7 0

Answer:

Explanation:

because amplitude of a sound wave is related to loudness of the sound, so the higher the amplitude, the louder the sound

hope this helps

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2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan

ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

$ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\ ac=0.0168m\\ac=16.8mm$

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.

5 0

3 years ago

Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water

artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x

density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0

3 years ago

You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __

SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy ${\rm KE}$ of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy ${\rm GPE}$ of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

5 0

1 year ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )