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3 years ago

The waves shown below represent sound waves. Which of the waves would have the highest-pitched sound?

Physics

1 answer:

Ostrovityanka [42]3 years ago

8 0

Humans hear frequencies from 20 Hz (low) up to 20,000 Hz (high)

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Which of the following wire gage sizes is the thickest? A. 8 B. 10 C. 14 D. 0

maks197457 [2]

A because it's the smaller the thicker you just can't have 0 gage

5 0

3 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

3 years ago

En que pais las mujeres pueden participar en los deportes ???

CaHeK987 [17]

im pretty sure it was tennis

3 0

3 years ago

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

6 0

3 years ago

A metal can containing condensed mushroom soup has mass 205 g, height 11.0 cm and diameter 6.38 cm. It is placed at rest on its

AlexFokin [52]

Answer: a) 0.000093 kg.m²

b) height of the can not required

Explanation:

It is placed at rest on its side, initial velocity = 0

s = ½ a t²

3 = ½ a 1.5²

a = 2.667 m/s²

torque T = f R = μ m g R cos Φ

according to Newton's second law of rotational motion,

T = I α = I (a/R) = μ m g R cos Φ

μ = I a/(R²m g cos Φ)

∑F = m a

m g sin Φ - μ m g cos Φ = m a

canceled all m, and remember that μ = I a/(R²m g cos Φ)

g sin Φ - I a/(R²m g cos Φ) g cos Φ = a

g sin Φ - I a/(R²m) = a

9.8 sin 24° - I (2.667)/(0.0319² * 0.205) = 2.667

3.86 - I x 2.667/0.000208 = 2.667

3.86 - 12822 x I = 2.667 m

I = 3.86 - 2.667 / 12822 = 1.193 / 12822 = 0.000093 kg.m²

6 0

3 years ago