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2 years ago

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A cross-country runner accelerates from 4 m/s to 7 m/s over the course of 5 seconds. What is the runner's average rate of accele

ration? A. 0.3 m/s/s B. 0.6 m/s/s C. 1.4 m/s/s D. 2.2 m/s/s

1 answer:

fredd [130]2 years ago

4 0

B. 0.6 for show
hopefully this works lemme know

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An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati

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Answer:

t = 106π / 30*2.1

Explanation:

$w_{i} = 1.06*10^{2}$ => 106

=> 106 x 2π/60

=> 106/30π

∝ = -2.1 rad/sec²

$w_{f}$ => 0

$w_{f}$ = $w_{i}$ + ∝t

∴ ( $w_{f}$ - $w_{i}$ ) / ∝ = t

t = 106π / 30*2.1

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2 years ago

It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

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Read 2 more answers

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 225 kg and it

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To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.

From Newton's second law we understand that

$F= ma (\rightarrow$ Gravity at this case)

Where,

m = mass

a= acceleration

Also we know that

$\rho = \frac{m}{V} \Rightarrow m = \rho V$

Part A) The buoyant force acting on the balloon is given as

$F_b = ma$

As mass is equal to the density and Volume and acceleration equal to Gravity constant

$F_b = \rho V g$

$F_b = 1.2*323*9.8$

$F_b = 3798.5$

PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then

$F_{net} = F_b -W$

$F_{net} = F_b -(mg+\rho_H Vg)$

$F_{net} = 3798.5-(9.8*225*9.8*0.179*323)$

$F_{net} = 1030N$

PART C) The additional mass that can the balloon support in equilibrium is given as

$F_{net} = m' g$

$m' =\frac{F_{net}}{g}$

$m' = \frac{1030}{9.8}$

$m' = 105Kg$

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3 years ago