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Maru [420]
3 years ago
12

40 POINTS I’ve been stuck on this for hours please help :(

Physics
2 answers:
kondaur [170]3 years ago
8 0

  ..................................         uh ya

Norma-Jean [14]3 years ago
7 0

Answer:

Explanation:

Particle         Relative Mass       Charge               Location

Proton             1                            +                        Nucleus

Electron          0                            -                        Outside rings (not nucleus)

Neutron.         1                            No Charge        Nucleus

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PLEASE HELP ASAP :|
barxatty [35]
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Note: The first part of the answer is a single sentence. The problem says in a complete sentence, so just in case that you need only one sentence you can take the first part. If you can add in more than a sentence, you can put in more from the second paragraph. 
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3 years ago
Which three factors are used to calculate gravitational potential energy?
NeX [460]

Answer:

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2 years ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
Most of the elements (85%) in the periodic table are.
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Most of the elements are metals
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3 years ago
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A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u
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Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

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3 years ago
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