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shtirl [24]
2 years ago
13

What is a situation where you can observe both kinetic and potential energy, shifting from one form to the other?

Physics
1 answer:
Novay_Z [31]2 years ago
6 0

Answer:

when an object kept at some height and having some mass then ,when it falls down then shifting of energy takes place

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Describe two ways unbalanced forces help you in your day to<br> day life.
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Answer:

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  2. we need unbalance force to drag objects
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3 years ago
a concave mirror has a focal length of 18 cm. where will an image form if an object is placed 58 cm from the mirror
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Answer:

here

Explanation:

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3 years ago
How much time will elapse between seeing and hearing an event?
julia-pushkina [17]

Depends on how far away the event is and what the temperature is as this affects the speed of sound.

For example, let's say you're 600 meters away and the temperature has no affect.

The speed of sound would be roughly 340 m/s so the time it would take to hear the sound would be 600/340 = 1.76 seconds

The speed of light (c) is 3.0 X 10^8 m/s so the time it would take to see the event would be 600/3 X 10^8 = 2 X 10^-7

Subtract: 1.76 - (2 X 10^-7) = approx. 1.76

3 0
3 years ago
What subatomic particle acts like a mini-magnet?
Semenov [28]
 The subatomic particles that acts like a mini-magnet is electron. Electrons are negatively charged sub atomic particles in an atom. The electron spin is a property of an electron that makes it behave like it's spinning; a spinning electron produces a magnetic field that makes it behave like a tiny magnet in an atom. 
5 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
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