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3 years ago

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A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in

thermal contact. The specific heat capacity of aluminum is more than double that of copper. Which object experiences the greater temperature change during the time the system takes to reach thermal equilibrium?

1 answer:

butalik [34]3 years ago

4 0

Answer:

copper will have more change in temperature as compare with aluminum

Explanation:

Hot piece of copper is made in contact with cold piece of aluminium

So here thermal energy transfer will take place from copper to aluminium

so by energy conservation we can say that heat given by copper is same as the heat absorbed by aluminium.

now we have

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

here we know that

$s_1$ = specific heat capacity of copper

$s_2$ = specific heat capacity of aluminum

given that specific heat capacity of aluminium is more than double that of copper

so we can say

$s_2 = 2s_1$

so here if the mass of copper and aluminium is same then

$\Delta T_1 = 2 \Delta T_2$

so temperature change of copper is twice the temperature change of aluminium

So copper will have more change in temperature as compare with aluminum

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The hawk’s centripetal acceleration is 2.23 m/s²

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$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

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therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

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1 year ago

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the

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As per the question, the mass of meteorite [ m]= 50 kg

The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

$Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

$=\frac{1}{2}50kg*[1000\ m/s]^2$

$=\frac{1}{2}50* 10^{6}\ J$

$=25*10^6\ J$

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$

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3 years ago

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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

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Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

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Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28 × 10⁵km

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T = 28.9 × 24 × 60 × 60

= 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

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3 years ago

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Answer:

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Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 40 m/s

Height (h) = 50 m

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Horizontal distance (s) =?

Next, we shall determine the time taken for the package to get to the ground.

This can be obtained as follow:

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

50 = ½ × 9.8 × t²

50 = 4.9 × t²

Divide both side by 4.9

t² = 50 / 4.9

t² = 10.2

Take the square root of both side

t = √10.2

t = 3.2 s

Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:

Horizontal velocity (u) = 40 m/s

Time (t) = 3.2 s

Horizontal distance (s) =?

s = ut

s = 40 × 3.2

s = 128 m

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