A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (

Afina-wow [57]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is

hypotenuse/ $\sqrt{2}$

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/ $\sqrt{2}$ km/h= 63,6396 km/h.

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

$\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h$ ,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= $sin^{-1}'frac{63,6396}{629,5851}$ =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get

Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows see fig 2

4 0

3 years ago

What distance is required for a train to stop if its initial velocity is 23 m/s and its

Varvara68 [4.7K]

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec) = 1,058 meters .

7 0

3 years ago

In what way are mercury and venus similar ?

Neporo4naja [7]

Answer:

they are both planets they are both made of rock and metel

Explanation:

3 0

3 years ago

Read 2 more answers

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman,

lina2011 [118]

Answer:

44.1 m

Explanation:

Given:

$V_a$ = speed of sound in air = 343 m/s
$V_r$ = speed of sound in the rod = $15V_a$
$\Delta t$ = times interval between the hearing the sound twice = 0.12 s

Assumptions:

$l$ = length of the rod
$t$ = time taken by the sound to travel through the rod
$T$ = time taken by the sound to travel to through air to the same point = $t+\Delta t = t+0.12\ s$

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

$l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}$ ..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

$l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m$

Hence, the length of the rod is 44.1 m.

4 0

3 years ago

1.05 MEASURING PHYSICAL PROPERTIES LAB REPORT

Nana76 [90]

Answer:

Objective(s): In your own words, what was the purpose of this lab?

The purpose of this lab is to get us familiar with the physical properties of different types of materials that are used for building.

Hypothesis: In this section, include the if/then statements you developed during your lab activity. These statements reflect your predicted outcomes for the experiment.

If we use clay brick to build the roof of the house, then the temperature inside the house will remain cooler.

If we use wood to build the walls and floors of the house, then the temperature inside the house will remain cooler.

If we use nickel to bring electricity into the home, then it will allow electricity to flow into the home at a faster rate.

If we use iron to construct the latches on the windows and doors, then the magnetism will keep the latches secure.

Procedure: The materials and procedures are listed in your virtual lab. You do not need to repeat them here. However, you should note if you experienced any errors or other factors that might affect your outcome.

Using the summary questions, clearly define the dependent and independent variables of the experiment.

Data: Record the results of each of your physical property tests in the table below.

Wood _ 3/10 _0 _0.12 W/(m·K) _0 S/m _ 4 g/cm3

Clay Brick _ 2/10 _1 _0.6 W/(m·K) _ 0 S/m _ 5.88235 g/cm3

Iron _ 4/10 _10 _80 W/(m·K) _1x107 S/m _9.09091 g/cm3

Aluminum_ 7/10 _ 0 _235 W/(m·K) _ 3.8x107 S/m _ 6.66667 g/cm3

Copper _ 6/10 _0 _400 W/(m·K) _3.8x107 S/m _9.52381 g/cm3

Nickel _ 5/10 _7 _ 91 W/(m·K) _1.4x107 S/m _9.43396 g/cm3

Conclusion: Your conclusion will include a summary of the lab results and an interpretation of the results. Please answer all questions in complete sentences using your own words.

Using two to three sentences, summarize what you investigated and observed in this lab.

In this, I investigated the effect that different types of physical properties had on six different materials. I also investigated which materials would be the best to build a house.

2. What building material did you use to build your house? Did your results support or fail to support your hypotheses?

For the roof of my house, I used clay bricks. For the walls and floors of the house, I used wood. To bring electricity into the home I used nickel. To construct the latches on the windows and doors, I used iron. My results supported my hypotheses because clay bricks and wood had the lowest thermal conductivity rates, nickel had the second highest electricity conductivity rate, and iron had the highest magnetism rate.

3. What were the densities of the materials you chose for the walls and floor for the home in Tiny World? Why do you think a building material's density is important when building homes or architectural structures?

The density of the material I chose for the walls and floors for the home in Tiny World was 4 g/cm3 (Wood). I think a buildings material density is important when building homes or architectural structures because in order for that structure to stay stable the materials that are used must be dense. If the materials are not dense, the structure would most likely collapse if it is placed under too much pressure.

4. Why wouldn’t you choose wood or aluminum for the latches on your house?

I wouldn’t choose wood or aluminum for the latches on my house because neither of those two materials are magnetic.

5. Which material has the highest thermal conductivity? Which material has the highest electrical conductivity? Explain why thermal and electrical conductivity is so high with this material.

Copper has the highest thermal conductivity. Aluminum and copper have the highest electrical conductivity. Thermal conductivity is high in copper because copper is a very thin type of material, and this allows heat to pass through it easily. Electric Conductivity is high in both aluminum and copper because they are both a very thin type of metal and because they are both so thin, electricity will pass through them quickly.

I'm sorry if any of the answers are wrong, this was just assigned to me and my teacher hasn't graded it yet.

4 0

3 years ago