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3 years ago

How does specific heat affect the rate at which an object changes temperature

2 answers:

OverLord2011 [107]3 years ago

6 0

<span>How does specific heat affect the rate at which an object changes temperature </span>All else being equal,* rate of heat transfer is proportional to the square of the temp difference between source and sink. Given a step change in input, heat transfer will change abruptly; then the rate of transfer will gradually decrease as source and sink temperatures approach each other. Ideally the rate of transfer will go asymptotically to zero as the source and sink attain equal temperatures. In a real system it will go to some stable minimum as heat leaks in/out in various directions.

<span>Since a sink with lower thermal capacity (specific heat x mass) will require less heat input for a given temperature change, heat transfer in such a system will drop off more quickly than it would in one with higher thermal capacity -- the reverse of what you're proposing. At the same time, the system will reach steady-state faster because there's less total heat in the system. </span>

Once a system has stabilized in steady-state, thermal capacity is no longer an issue.

stira [4]3 years ago

5 0

If it is lower, it heats faster, if it's higher, it takes longer to heat.

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How should the student change the circuit to give negative values for current and

natulia [17]

Answer:

changing the polarity or direction of the battery changes the sign of the voltage and the current

Explanation:

The sign of current and voltage are due to established conventions.

The way that a DC circuit with negative current values is by changing the polarity of the power source or by inverting the battery, this creates that the electrons move in the opposite direction

Changing the battery also changes the direction of the power difference, since the potential from positive to negative, in most cases negative is assigned a potential of zero volts

In summary, changing the polarity or direction of the battery changes the sign of the voltage and the current

3 0

2 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

2 years ago

Nancys airplane trip took 4 hour. For one-fourth of that time, the airplane flew at a speed of 880 km/h, it flew at a speed of 6

9966 [12]

So for a minute lets ignore the 880 km/h. If it took 4 hours and she flew at 600 km/h 600*4=2400. Now lets Look at the 880 bit. If it took 4 hours and she where to fly at 880 it would've been 880*4=3520. Lets do 2400-600=1800, now we've got the 600 kmh bit done. Now lets see if you fly 880 km/h for one hour then you add 1800+880=2680.

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4 0

2 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$