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3 years ago

QUESTION 1 If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy?

2 answers:

7 0

1. U = Q + W
U = -500 + 1000
U = 500 J

2. The first law of thermodynamic is about the law of conservation of energy where energy in should be equal to energy out.

3. It is the windmill that does not transform energy from heat to mechanical instead it is the transforms the opposite.

4. In a heat engine, work is used to transfer thermal energy from a hot reservoir to a cold one.

5. 5.00 × 10^4 J - 2.00 × 10^4 J = 3.00 × 10^4 J

6. To increase the work done, we raise the temperature of the cold reservoir.

denpristay [2]3 years ago

4 0

Sallyablow has them all correct except 4 and 6.

4. D
6. C

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Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

4 years ago

A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th

Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2 V² = g x 2r + 1/2 v²

V² = g x 4r + v²

V² = 9.8 x 4 + 8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T = mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0

4 years ago

Object approaching or receding, and at what speed?

julia-pushkina [17]

Answer:

where is the figure...

5 0

3 years ago

What is Angular acceleration, please explain this concept. Give any equations that involve angular acceleration and explain them

Eva8 [605]

Answer:

Angular acceleration is defined as the rate of change of angular velocity of a body.

consider the attached figure as shown

It rotates with an angular velocity $\omega$

An point inside the object rotates along the path as indicated thus turning by an angle $\theta$ in time 't'

Thus we have

$\alpha =\frac{d\omega }{dt}\\\\=\frac{d}{dt}(\frac{d\theta }{dt})\\\\\therefore \alpha =\frac{d^{2}\theta }{dt^{2}}$

physically angular acceleration can be understood as the rate at which the angular speed of any object is changing with time.

8 0

3 years ago

What is the unit for IMA?

Dahasolnce [82]

It's unitless.

20 characters

7 0

3 years ago