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Vanyuwa [196]
3 years ago
8

Which words identify the scientific parts of a wave? Check all that apply.

Physics
1 answer:
uysha [10]3 years ago
6 0

Answer:

the question is not complete but some parts of a wave include crest and trough

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a cheerleader throws herself into the air with a velocity of 12 m/s at an angle of 75 degrees above the horizontal. what are the
NISA [10]

Answer:

Vx = 3.10 [m/s]

Vy = 11.59 [m/s]

Explanation:

To solve this problem we must decompose the velocity vector by means of the angle on the horizontal.

v = 12 [m/s]

Vx = 12*cos (75) = 3.10 [m/s]

Vy = 12*sin (75) = 11.59 [m/s]

7 0
3 years ago
The number of protons and neutrons in an atom is that atom’s _____________ number.
astraxan [27]

Answer:

The mass number of an atom is its total number of protons and neutrons. Atoms of different elements usually have different mass numbers , but they can be the same. For example, the mass number of argon atoms and calcium atoms can both be 40.

Explanation:

Please mark as brainliest

3 0
4 years ago
Read 2 more answers
Which isotope has 16 neutrons and is used to study nucleotides and nucleic acids?
qaws [65]
When naming isotopes, the number beside it indicates the mass number. Now mass number is the sum of protons and neutrons. 

We know that the mass number of a stable Sulfur atom is 32, so  we can rule out C as an isotope. The atomic number of Sulfur is 16 and that means that there are 16 protons. Now if we subtract 16 from 33, then that means there are 17 neutrons. So we can rule out D. 

Phosphorus on the other hand has an atomic number of 15. If we subtract 15 from 31 we will have 16. That means that Phosphorus-31 has 16 neutrons.

The answer would then be A. Phosphorus 31. 
7 0
3 years ago
Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m
Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

A=\frac{1-V_{1}^{2}  }{E_{1} }-\frac{1-V_{2}^{2}  }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}

a) The maximum contact pressure is:

P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} })  }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa

The width of contact is:

b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} })  } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0
3 years ago
An object of mass 2 kg travels through outer space in a straight line at a constant
slava [35]

Answer:

b. It is subject to a net force of zero

Explanation:

F = m*a

F = force

m = mass

a = acceleration

Since the object is moving at a constant velocity not speeding up or slowing down and moving in a sight line. All the forces that are acting on the object are canceling each other out so the answer is:

b. It is subject to a net force of zero

7 0
3 years ago
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