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3 years ago

Two objects made of the same material are heated to 60 oC and 90 oC. According to

Physics

2 answers:

earnstyle [38]3 years ago

6 0

Answer:

The newton law of cooling is:

dQ/dt = -h*a(To - Ta)

Where h and a are constants, To is the temperature of the object and Ta is the temperature of the ambient.

this means that initially if in both cases Ta is the same, the object with higher temperature cools faster.

Now, obviously, if we want to know which object cools faster to a given temperature, for example, 30°C, the object with 60°C will reach that temperature in less time than the other one. (this is because the hot object cools faster at the beginning, but as it starts to lose temperature, the rate at which it colls starts to decrease)

nlexa [21]3 years ago

4 0

Assuming that there is in a vacuum, the two object will cool at the same rate, because the objects are made of the same material they will have the same cooling rate, assuming the surrounding temperature is the same.

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An infinite plane of charge has surface charge density 7.2 Î¼C/m^2. How far apart are the equipotential surfaces whose potential

Rina8888 [55]

Answer:

so the distance between two points are

$d = 0.246 \times 10^{-3} m$

Explanation:

Surface charge density of the charged plane is given as

$\sigma = 7.2 \mu C/m^2$

now we have electric field due to charged planed is given as

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now we have

$E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}$

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now for the potential difference of 100 Volts we can have the relation as

$E.d = \Delta V$

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$d = \frac{100}{4.07 \times 10^5}$

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3 years ago

A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the

Mrac [35]

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with $u=5.4\ m/s$

Height of helicopter $h=105\ m$

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

$\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}$