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2 years ago

10

You are stranded in a blizzard. You need water to drink to drink,and you're trying to stay warm.should the melt the snow and dri

nk it or just eat the snow? Explain

1 answer:

Crazy boy [7]2 years ago

7 0

It would be a really bad idea to eat the snow because you obviously are trying to stay warm right? Well, the best thing to do is melt the snow. However, the process of melting the snow would have a few complications as well. But yes, the latter idea (drinking the snow) is a better idea (not the best).

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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

1 year ago

A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only

ANTONII [103]

Answer:

Technician A is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0

2 years ago

If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?

kicyunya [14]

For this case we have that by definition:

$F = ma$

Where,

<em>m: mass of the object </em>
<em>a: acceleration of the object </em>
<em>F: summation of forces </em>

We have then:

$F = 10 + 5\\F = 15 N$

Then, by clearing the acceleration we have:

$a = \frac {F} {m}$

Substituting values we have:

$a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}$

Answer:

The acceleration of the box is equal to:

$a = 3 \frac {m} {s ^ 2}$

6 0

1 year ago

A snowball is melting at a rate of 324π mm3/s. At what rate is the radius decreasing when the volume of the snowball is 972π mm3

Oduvanchick [21]

Answer:

The radius is decreasing at 4 mm/s

Explanation:

The volume of a sphere is:

$V = 4/3*\pi *r^3$ So, when the volume is 972π mm^3 the radius r is:

r = 9mm

Now, the change rate is given by the derivative:

$dV/dt = 4/3*\pi *3*r^2*dr/dt$

Where: dV/dt = -324π mm^2/s

r = 9mm

Solving for dr/dt:

dr/dt = -4mm/s

5 0

1 year ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

1 year ago