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3 years ago

Why did scientist's gather at Los Alamos in 1945.

1 answer:

7 0

Los Alamos was the site for the Trinity Test, which was performed as a part of The Manhattan Project. They tested nuclear bombs to ensure it worked the way they intended it to.

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Which type of mirror causes light to spread out?

saw5 [17]

A convex mirror makes a reflected light rays spread out.

3 0

4 years ago

Read 2 more answers

A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the

ANTONII [103]

Either A or D but maybe D

4 0

3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago

If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in

Ksenya-84 [330]

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

6 0

3 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 83 vol% carbide

velikii [3]

Answer:

k_max = 31.82 w/mk

k_min = 17.70 w/mk

Explanation:

a) the maximum thermal conductivity is given as

$k_{max} = k_m*v_m + k_p*v_p$

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.17

v_p = proportion of carbide in the cement = 0.83

= 66*0.17 + 28*0.83

k_max = 31.82 w/mk

b) the minimum thermal conductivity is given as

$k_{min} = \frac{k_{carbide}* k_{metal}}{k_{metal} v_{carbide} +k_{metal} v_{carbide}}$

= \frac{28+66}{28*0.17 +66*0.83}

k_min = 17.70 w/mk

7 0

3 years ago