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4 years ago

8

The density of the object on the left is 1.5 g/cm3 and the density of the fluid is 1.0 g/cm3. Which has greater density? The obj

ect will in the water column.

2 answers:

gizmo_the_mogwai [7]4 years ago

4 0

the first one is object and the second one is sink

LiRa [457]4 years ago

4 0

Answer:

object, sink

Explanation:

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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

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To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

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Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t

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1. The bowling ball have more potential energy as it sit on top of the building

2.The bowling ball have same potential energy and kinetic energy as it is half way through its fall

3. The bowling ball have more kinetic energy just before it hits the ground

4. The potential energy of the bowling ball as it sits on top of the building is 784J

5. The potential energy of the ball as it is half way through the fall, 20 meters high is 392J

6. The kinetic energy of the ball as it is half way through the fall is 392J

7. The kinetic energy of the ball just before it hits the ground is 784J

Explanation:

Calculating potential energy and kinetic energy for all the instances,

1. ball on top of a 40 meters tall building

Potential energy at the top of building with a height of 40m = mgh

P.E = mgh =2*9.8*40= 784J

At the top pf the building since v=0 kinetic energy is zero

2. half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second

Potential energy when it is half way through fall = mgH

where H represents new height that is equal to 20m

hence P.E=mgH=2*9.8*20= 392J

Kinetic energy of the ball is $\frac{1}{2} mv^{2} = \frac{1}{2} *2*19.8^{2}$ =392.04J

3. Just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.

The potential energy of ball just before it hits the ground = mgh= 2*9.8*0=0J

kinetic energy = $\frac{1}{2} mv^{2}$ = 784J

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3 years ago