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pashok25 [27]
3 years ago
10

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

8 0
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
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