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4 years ago

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

Read 2 more answers

A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is

evablogger [386]

Answer:

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