6 مسائل تنتظر ابداعاتكم "ممكن مساعده"

[الخاطري]

السلام عليكم ورحمة الله وبركاته
لوسمحتم ممكن تساعدوني بحل هالمسائل
ابغاها قبل الاحد
وراح اكون ممنونه لكم


Ex1: A tow truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 0.50 cm . when the car just begins to move , the tension in the cable is 890 n . how much has the cable stretched? (y= 2.0 x10¹¹ N/m²)



Ex2: A solid , square pinewood raft measures 4.0 m on a side and is 0.30 m thick .
(A ) determine whether the raft floats in water, and (b) if so , how much of the raft is beneath the surface . (the pine wood density = 550 kg/m³)(Archimedes principle)




Ex3: A time and temperature sign on a bank indicates that the outdoor temperature is
-20 ºC . find the corresponding temperature on the Fahrenheit scale.


Ex4: A temperature of absolute zero occurs at -273.15 ºC . what is the temperature on the Fahrenheit scale?


Ex5: A lead section of the Alaskan pipeline had a length of 65m and a temperature of 18 ºC when it was installed. What is its
Change in length when the temperature drops to a frigid -45 ºC ?
(α lead = 29 x 10 ¯6)


Ex6: A 125 g block of an unknown substance with a temperature of 90.0 ºC is placed in a Styrofoam cup containing 0.326 kg of water at 20.0 ºC . the system reaches an equilibrium temperature of
22.4 º C . what is the specific heat , Cx , of the unknown substance if the heat capacity of the cup is neglected .

[الخاطري]

تكفون اللي يعرف يساعدني

عشان يمديني اذاكر..


:mommy_cut::mommy_cut:

[فطوومة]

الخاطري

هاي اسئلة الشيت الثاني ..هههههه

السؤال الاول حليته في موضوعي .. بس ناقص

واتلثاني حليته .. روح اكتبه وارجع .. بالتوفيق

[الخاطري]

هههههههههه
وشلون عرفتي شكلك
معنا بالجامعه
وحاسه بمأساتي :a_plain111:

الله يعينا بس
ومشكوره مره

[د.امير]

Q1
A tow truck is pulling a car out of a ditch by means of a steel cable that has the length ( Lo) = 9.1 m
The radius ( r) = 0.50 cm.
= 0.005m
Then the area of the cross section is (A) = πr2
= (3.14)(0.005m)2
= 7.85*10-5m2
When the car just begins to move , the tension in the cable is (F) = 890 N
The young's modulus of the Y steel = 2.0 x 10¹¹ N/m²
We know that magnitude of F is given by
F = Y (ΔL/Lo)A
Then the stretched length in the cable is give by
Delta L = FL0/YA
= ( 890 N)(9.1m)/( 2.0 x 10¹¹ N/m²)(7.85*10-5m2)
= 8099 /15.7*106
= 515.859*10-6m
= 0.516mm

[د.امير]

Q2
We must figure out the raft’s weight and the buoyant force to stabilise the weight:
Given:
The density, of pine = 550 kg/m3
The volume of the wood = 4.0m*4.0m*0.30m=4.8m3 . Hence,
Weight of raft= (550 kg/m3)(4.8m3)(9.80 N/kg)=26 000 N.
When the raft is below surface level displacing 4.8 m3 of water, the maximum buoyant force exists and is equal to the weight of displaced water. Along with water’s density, it can be determined.
FB=Wfluid =(1000 kg/m3)(4.8 m3)(9.80 N/kg) = 47 000N
Since the maximum buoyant force is larger than the raft’s weight, then the raft will be afloat (Answer A).
B) the volume of the raft that will be beneath the water surface when
FB= Weight of raft = 26 000 N, or
FB= (1000 kg/m3) V (9.80 N/kg) = 26 000N, or
V = 2.65 m3
2.65 m3 = 4.0m*4.0m* t, where t is the depth of wood that will be beneath water,
t= 0.17m

[د.امير]

Ex3: A time and temperature sign on a bank indicates that the outdoor temperature is
-20 ºC . find the corresponding temperature on the Fahrenheit scale.


This is direct change from degrees centigrade to Fahrenheit,
F = 1.8 C + 32
F = 1.8 (-20) + 32 = - 4

Ex4: A temperature of absolute zero occurs at -273.15 ºC . what is the temperature on the Fahrenheit scale

F = 1.8 C + 32
F = 1.8 (-273.15) + 32 = - 459.7

[د.امير]

Q5
solid materials expand upon heating and contract when cooled. The change in length is expressed by:
ΔL = L0 α ΔT
where
L0 original length,
α = thermal coefficient of expansion
and
ΔT is change in temperature

ΔL= 65 * 29 10 -6 * (-45 -18) = -0.119 m (contraction

[د.امير]

Q6
Q is the quantity of heat, m is the mass of the body, Cx is the specific heat and T the temperature change. or
Q = m Cx ΔT
Heat lost by the unknown material = 0.125 kg Cx (22.4 – 90)
Heat gained by water = 0.326 kg * 4186 ( 22.4 – 20), specific heat of water is 4186 J kg-1 K-1
Assuming equilibrium, then the heat lost by the unknown material is the heat gained by water, or
Qunknown + Qwater =0
0.125 kg Cx (22.4 – 90) + 0.326 kg * 4186 ( 22.4 – 20) =0
8.45 Cx = 3275.1, or Cx= 387.6 J kg-1 K-1 specific heat of the unknown material

[الخاطري]

د:اميـــر
ماادري كيف اشكرك
على مجهودك
لك خالص شكري
تحياتي

[فطوومة]

سو سوري

انشغلت ..

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