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3 years ago

13

Consider a bicycle that has wheels with a circumference of 2 m. Solve for the linear speed of the bicycle when it's wheels rotat

e at 1 revolution per second

1 answer:

Yuliya22 [10]3 years ago

4 0

Answer:0.3183m/s

Explanation: V=wr

V=linear velocity

W= angular velocity

P=2πr

r=p/2π

r=2/2π

r= 1/3.142

r=0.3183

V= 0.3183*1

V=0.3183m/s

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Answer:

The ball will fall back and land to Elle's hands.

Explanation:

The bus move in a straight line with constant velocity means that there is no change of direction and no acceleration. Inertia can change the direction of the ball and acceleration can change its velocity. Since these two factors is not present in this scenario, the ball only has vertical movement. Thus the ball will land where it was thrown, in Elle's hands.

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A submerged submarine alters its _____ to rise or fall in the water.

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4 years ago

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You carry a 5 kg sack of potatoes across the store for 5 minutes, and you wait in line holding it for 30 seconds before you get

ludmilkaskok [199]

Answer:

The only work done is when the person lifts the sack over a distance, W = 78.48 [N]

Explanation:

We have to remember the definition of work, which tells us that work is the result of a force by a distance, we must apply this concept in each of the movements of the person in the problem described.

W = F * d

where:

F = force [N]

d = distance [m]

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3 years ago

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is

AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

$W'=W-B$

which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

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

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

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3 years ago