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3 years ago

The term _______ is used to indicate the frequency level of a sound.

Physics

2 answers:

SOVA2 [1]3 years ago

7 0

the term WAVELENGTH is used to indicate the frequency level of a sound

a wavelength is the correct answer

aksik [14]3 years ago

6 0

The answer is D, pitch

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The gasoline in a car does 40,000 J of work on a car and generates a constant force of 20 N. How far did the car go?

AnnyKZ [126]

L=F•d=>d=L/F=40,000/20=2,000 m

7 0

2 years ago

What is the kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18 m/s? A.450 J B.900 J C.8,100 J D.16,000

Svet_ta [14]

KE = (1/2)·(mass)·(speed)²

KE = (1/2)·(50 kg)·(18 m/s)²

KE = (25 kg)·(324 m²/s²)

KE = 8,100 kg-m²/s²

KE = 8,100 Joules

5 0

3 years ago

Read 2 more answers

WILL UPVOTE!!!Physics help please!!

liubo4ka [24]

Speed v = initial speed u + acceleration a x time t
v=u+at = 2 + 4*3 = 14 m/s

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3 years ago

Humans evolved in Earth's atmosphere, therefore the pressures interior to the body are relatively close to atmospheric pressure.

harina [27]

Answer:

194516 sheets

Explanation:

So the area of each sheet of paper is:

A = 0.216 * 0.279 = 0.060264 square meters

For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be

F = AP = 0.060264 * 101325 = 6106 N

Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is

m = F/g = 6106 / 9.81 = 622.4 kg

If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is

622.4 / 0.0032 = 194516 sheets

4 0

3 years ago

Read 2 more answers

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$

Now

$F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf$

$F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf$

Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

The stress in the steel is $21678.47223\ lbf-in^2$

$\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2$

The stress in the steel is $383.1109\ lbf-in^2$

4 0

3 years ago