Which quantity must be the same for two bodies in thermal equilibrium?

A 2.5 g bullet traveling at 350 m/s hits a tree and slows uniformly to a stop while penetrating a distance of 12 cm into the tre

Cloud [144]

Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

V = velocity of the body = 350m/s

Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

Work done is the force applied to body to cause it to move through a distance.

Work = Force × distance

Force = ma = 0.0025 × 10

Force = 0.025N

Distance = 12cm = 0.12m

Work = 0.025×0.12

Work = 0.003Nm

work done by the tree in stopping the bullet is 0.003N

4 0

3 years ago

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Where does a phase change occur

il63 [147K]

A phase change is occuring; the liquid water is changing to gaseous water, or steam. On a molecular level, the intermolecular forces between the water molecules are decreasing. The heat is providing enough energy for the water molecules to overcome these attractive forces.

3 0

2 years ago

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is: