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3 years ago

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A weightlifter lifts a 125-kg barbell straight up 1.15 m in 2.5 s. What was the power expended by the weightlifter?

1 answer:

labwork [276]3 years ago

7 0

Answer:

560

Explanation:

Round your answer to two significant figures.

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A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$

7 0

4 years ago

Which term describes the amount of charge that passes a point in a circuit each second?

mestny [16]

Current, I got it right on my quiz

5 0

2 years ago

Consider the elementary gas-phase reversible reaction A 3C Pure A enters at a temperature of 400 K and a pressure of 10 atm. At

xxMikexx [17]

Answer:

39%

Explanation:

The equilibrium equation is:

$A\rightleftharpoons 3C$

The initial concentration of A can be calculated from the ideal gas equation:

$pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter$

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

A ⇄ 3C

I 0.304 0

C - x + 3x

E 0.304 - x 3x

$K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}$

Solve for x:

You need to use a graphing calculator:

108x³ = 0.304 - x

108x³ + x - 0.304 = 0

x ≈ 0.1195 mol/liter

Then:

$C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter$

The equilbrium conversion is:

$\% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100$

$\% \approx 39\%$

3 0

3 years ago

Please help fast!! I need this in less than 17 hours!

djyliett [7]

SOLUTION is given in attachment below.

7 0

3 years ago

A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy,

Cerrena [4.2K]

Answer:

let M be the mass of brick

let m be the mass of the pebble

.5M(3)^2 =KE

.5m(5)^2= KE

.5M9 = .5m25

9M = 25m

(9/25)M = m

7 0

3 years ago