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3 years ago

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How much heat is required to heat 0.1 g of ∆hvap =2260 j/g ∆h =340j/g fus iceat−30ctosteamat100c? use the approximate values?

1 answer:

Anestetic [448]3 years ago

7 0

How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>

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A person consumes a snack containing 14 food calories (14kcal). what is the power this food produces if it is to be "burned off"

Inessa05 [86]

Answer:

B) 2.7W

Explanation:

Converting Cal to Joule

1 cal = 4.186J

14 kcal = 14 x 1000 x 4.186

= 58604 J

Converting hour to seconds

6 hours = 6 x 60 x 60 seconds

= 21600 seconds

Power is the time rate of doing work.

Power = Work/Time

P = (58604) / (21600)

P = 2.7W

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3 years ago

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

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3 years ago

An electroscope is a simple device consisting of a metal ball that is attached by a conductor to two thin leaves of metal foil p

baherus [9]

Answer:

the electroscope separate by the presence of charge carriers

Explanation:

Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where

Fe - Tx = 0

Fe = Tx

In summary, the electroscope separate its leaves by the presence of charge carriers

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3 years ago

A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg

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1 m/s^2

Using F=ma,

7=7a

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3 years ago

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Zak, helping his mother rearrange the furniture in their living room, moves a 60 kg sofa 5.9 m with a constant force of 40 N. Wh

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Answer:236 J

Explanation:

Given

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Displacement $s=5.9 m$

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neglecting Friction

Work done is given by

$W=F\cdot s$

$W=F\cdot s\cos 0$

$W=40\cdot 5.9$

$W=236 J$

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3 years ago

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