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qwelly [4]
3 years ago
14

What happens to the volume of a loaf of bread that is squeezed? The mass? The density?

Physics
1 answer:
nalin [4]3 years ago
3 0

Explanation:

The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.

I hope this helped!

Thanks!

Your friend in answering,

~Steve

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A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals
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Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

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Q = It............... Equation 3

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t = Q/I.................. Equation 4

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5 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

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To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

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\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

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W=F_f d cos \theta

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cos \theta=0

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(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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