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pentagon [3]
4 years ago
12

Julian tests four springs and records the results in the table

Physics
2 answers:
Tju [1.3M]4 years ago
8 0

Answer:

list of spring in order of increasing spring constant is given by,

W, Y, X, Z

Explanation:

Spring W:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 20 / 0.5

k = 40 N/m

Spring X:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 60 / 0.3

k = 200 N/m

Spring Y:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 40 / 0.4

k = 100 N/m

Spring Z:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 50 / 0.1

k = 500 N/m

Thus, list of spring in order of increasing spring constant is given by,

W, Y, X, Z





Cerrena [4.2K]4 years ago
4 0

the correct answer is a. w,y,x,z.

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Answer:

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Explanation:

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The highest barrier that a projectile can clear is 19.7 m, when the projectile is launched at an angle of 15.0o above the horizo
svlad2 [7]

Answer:

76 m /s

Explanation:

maximum height, H = 19.7 m

Angle of projection, θ = 15 degree

Let u be the projectile launch speed

Use the formula of maximum height

H = u^{2}Sin\theta ^{2}/2g

19.7 = u^{2}Sin^{2}15/19.6

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1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
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Predict the date of the next full moon.
lapo4ka [179]

Answer:

The 19th

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A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

C = \frac{855.6}{220^2 \times 2\pi \times 60}

C = 4..689 \times 10^{-5} Faraday

C = 46.891 \mu F

6 0
4 years ago
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