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zhenek [66]
3 years ago
10

The sprinter experiences ___________ from A to B to C.

Physics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

The sprinter experiences muscle cramps and a stray dog on the track from A to B, followed immediately by acute nausea and a gaggle of news photographers from B to C.

If none of these selections appears on the list of choices, then perhaps we can do better if you'll permit us a glance at the picture that accompanies the question, even if you still don't let us see the choices.

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Tatiana [17]
<span>Crust. The thin solid outermost layer of Earth. ...Asthenosphere. The lower layer of the crust. ...Lithosphere.Plasticity: is solid but still being able to. flow without being a liquid.The cool, rigid outermost layer of the Earth. ...<span>the solid part of the earth consisting of the crust and outer mantle.</span></span>
6 0
3 years ago
Determine if the data are qualitative or quantitative.
Amanda [17]

Answer:

Qualitative, Quantitative, Qualitative, Quantitative, and Qualitative.

Explanation:

7 0
2 years ago
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Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typicall
joja [24]

Answer:

a) N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) diameter of single rope with the same effect = 2.415 cm

Explanation:

Hooke's law explains that stress is directly proportional to strain.

Stress ∝ Strain.

Stress = E × Strain

E = constant of proportionality = Young's Modulus = 4.0 ✕ 10⁹ N/m².

Stress = (Load/Total Cross sectional Area)

Load = a pair of 85 kg mountain climbers = 85 × 2 × 9.8 = 1666 N

Total Cross sectional Area = (Number of strands) × (Area of one strand) = A

Strain = (ΔL/L)

ΔL = 1.00 cm = 0.01 m

L = 11 m

Strain = (0.01/11) = 0.0009091

Stress = (Young's Modulus) × (Strain) = (4.0 ✕ 10⁹) × (0.0009091) = 3,636,363.64 N/m²

(Load/ total Area) = 3,636,363.64

Total area = (Load/3,636,363.64) = (1666/3,636,363.64) = 0.00045815 m²

Recall,

Total Cross sectional Area = (Number of strands) × (Area of one strand)

Area of one strand = (πd²/4)

diameter of one strand = 2 μm = (2×10⁻⁶) m

Area of one strand = (πd²/4)

= π × (2×10⁻⁶)² ÷ 4 = (3.142 × 10⁻¹²) m²

Total Cross sectional Area = (Number of strands) × (Area of one strand)

0.00045815 = N × (3.142 × 10⁻¹²)

N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) If it was a single rope, the cross sectional Area would just be equal to the total cross sectional Area obtained in (a)

A = 0.00045815 m²

A = (πD²/4)

where D = diameter of the single rope

0.00045815 = (πD²/4)

D² = (4×0.00045815) ÷ π = 0.0005833347

D = 0.02415 m = 2.415 cm

Hope this Helps!!!

6 0
3 years ago
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A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10
anzhelika [568]

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by  Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90°  = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.

5 0
3 years ago
Is the frictional force the same as the applied force when the net force equals zero?
DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

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