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dedylja [7]
4 years ago
6

What effects occur when heat energy is added to a system

Physics
1 answer:
Kobotan [32]4 years ago
6 0
It could possibly melt things
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An example of a constant acceleration is
serious [3.7K]
An example is free fall ,
8 0
3 years ago
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A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is
evablogger [386]

Answer:

5m

Explanation:

Using Pythagoras theorem,

a^2+ b^2=c^2

3^2+4^2=c^2

25=c^2

√(25)=c

5m=c

6 0
3 years ago
An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12
kati45 [8]

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} +  \frac{1}{q}

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

     

we calculate

          \frac{1}{q} = - \frac{1}{12} - \frac{1}{29}

          \frac{1}{q} = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          m = \frac{h'}{h} = - \frac{q}{p}

       

we substitute

          m = - \frac{-8.488}{29}

          m = 0.29

the positive sign indicates that the image is right

4 0
3 years ago
How is 0.00069 written in scienitfic notation?
victus00 [196]
You need to move the decimal point between the six and nine. 6.9 X 10^-4
8 0
3 years ago
A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
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