[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

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