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bagirrra123 [75]
3 years ago
6

An iron rod and a copper rod of equal length are each held by hand at one end, with the other end in the flame from a Bunsen bur

ner. The copper rod becomes too hot to hold much sooner than the iron rod. What does this information tell you about iron and copper?
Physics
1 answer:
lara31 [8.8K]3 years ago
3 0

Answer:

Iron has a higher heat capacity than copper

Explanation:

The heat capacity of a substance is defined as the amount of heat required to raise the temperature of an object by 1°C.

The higher the heat capacity of a substance, the longer the time it takes for the substance to become hot when heated.

Since the copper rod becomes too hot to hold much sooner than the iron rod, it means that iron has a higher heat capacity than copper.

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A deuteron consists of one proton and one neutron. A deuteron moving horizontally enters a uniform, vertical magnetic field of 0
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Answer:

Explanation:

Let the velocity of deuteron be v then force on it in magnetic field

Bqv , B is magnetic field and q is charge on deuteron . This force will provide centripetal force for circular path so

mv² / r = Bqv   m is mass of deuteron and r is radius of circular path

v = Bqr / m

(.5 x 1.6 x 10⁻¹⁹ x 55.6 x 10⁻² )/ 3.34 x 10⁻²⁷

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A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are mov
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  <span>net work = change in kinetic energy 

for Block B, we just have the force from block A acting on it 
F(ab)d= .5(1)vf² - .5(1)(2²) 
F(ab)d= .5vf² - 2 

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction 

10-F(ba)d = .5(4)vf² - .5(4)(2²) 
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F(ba)d = 18 - 2vf² 

now we have two equations: 
F(ba)d = 18 - 2vf² 
F(ab)d= .5vf² - 2 

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab) 

10-.5vf² + 2 = 2vf² - 8 
12 - .5vf² = 2vf² - 8 
20 = 2.5vf² 
vf² = 8 

they both will have the same velocity 
KE of block A= .5(4)(2.828²) = 16 J 
KE of block B=.5(1)(2.828²) = 4 J</span>
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The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch
AnnZ [28]

Answer:

Explanation:

Given an LC circuit

Frequency of oscillation

f = 299 kHz = 299,000 Hz

AT t = 0 , the plate A has maximum positive charge

A. At t > 0, the plate again positive charge, the required time is

t =

t = 1 / f

t = 1 / 299,000

t = 0.00000334448 seconds

t = 3.34 × 10^-6 seconds

t = 3.34 μs

it will be maximum after integral cycle t' = 3.34•n μs

Where n = 1,2,3,4....

B. After every odd multiples of n, other plate will be maximum positive charge, at time equals

t" = ½(2n—1)•t

t'' = ½(2n—1) 3.34 μs

t" = (2n —1) 1.67 μs

where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

t''' = ½(2n—1) 1.67μs

t"' = (2n —1) 0.836 μs

where n = 1,2,3...

6 0
3 years ago
How does Borg’s Rating of Perceived Exertion relate to Target Heart Rate?
Nina [5.8K]

Answer:

it helps you estimate how hard you're working (your activity intensity). <u>perceived exertion</u> is how hard you think your body is exercising. ratings on the scale are related to heart rate (how hard your heart is working to move blood through your body).

Explanation:

7 0
3 years ago
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