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3 years ago

Hi!

Physics

1 answer:

dangina [55]3 years ago

4 0

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

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The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G

Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1

lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0

3 years ago

A 90kg person jumps from a 30m tower into a tub of water with a volume of 5m3 initially at 20°C. Assuming that all of the work d

Kryger [21]

Answer:

T₂ = 20.06 ° C

Explanation:

Given

P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C

Using the formula to determine the final temperature of the water

T₂ = T₁ * P * h / Eₐ * c

The work done of the person to the water

Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²

Eₐ = 49000 N

T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]

T₂ = 20.06 ° C

4 0

4 years ago

Earth is the only planet known to have _____.

postnew [5]

the answer is bioshere

4 0

3 years ago

Read 2 more answers

5. A cable is attached 32.0 m from the base of a flagpole that is about to

soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, $F_v$ = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

$F_v$ + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0

3 years ago