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4 years ago

7

michelle withdrew 120$ from her bank account. she now has 3345$ in her account qrite and solve an equation to find how much mone

y m was in her account before she made the withdrawal

2 answers:

Lerok [7]4 years ago

8 0

He had $3,465 before he with drew $120

Rama09 [41]4 years ago

5 0

3345+120=$3465 she had $3465 before the withdrawal

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3 years ago

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A wave is sent along the first rope transmitting a power of 57.3 W. It has a wavelength of 5.54 cm and velocity of 13.87 m/s The

Serhud [2]

Answer:

$A = 2.43*10^{-3} m$

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power through string can be determined as shown in figure

$P = 2\pi ^2 HVA^2F^2$

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V = 13.87 m/s

H = 567 g/m

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$V = f *\lambda$ $\lambda = \frac{v}{f}$

therefore $P = 2\pi ^2 HVA^2(\frac{v}{f})^2$

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3 years ago

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0

3 years ago