Download King Saud University 14 marks College of Science W= T× cos60 × d

King Saud University
College of Science
14 marks
Department of Physics and Astronomy
Second Term (1430-31)
Thursday 13 Jumada II 1431 h
1st mid-term
27 May 2010
Phys 145 (106)
10 30 a.m
Q1
A box is pulled on horizontal non friction surface by a rope that makes a 60° angle to the
horizontal direction. The tension in the rope is T = 70 N, and the box moves a distance d = 15
m. The work done in pulling the box is:
a. 288 J
b. 250 J
c. 125 J
d. 525 J
e. 500 J
0
W= T× cos60 × d = 70 × 0.5 ×15 = 525 J
Q2
A 1500 kg car moving at 10 m/s hits head-on a 3000 kg in rest. If the collision is completely
inelastic, the dissipated mechanical energy is
a. 25 kJ
K1 =
b. 100 kJ
c. 150 kJ
e. 50 kJ
1
1
m1v12 = × 1500 × 10 2 = 75000 J
2
2
m1v1 = (m1 + m2 )V ⇒ V =
K' =
d. 75 kJ
m1v1
=
(m1 + m2 )
m12 v12
1
1
1 m12 v12
1 (1500) 2 × 10 2
(m1 + m 2 )V 2 = (m1 + m2 )
=
=
×
= 25000 J
2
2
4500
(m1 + m2 ) 2 2 (m1 + m 2 ) 2
∆K = K 1 − K ' = 75000 − 25000 = 50000J = 50 kJ
Q3.
A pipe of radius “r” ends with a nozzle of radius “r/3”. If the average velocity of water
flowing in the pipe is v1, the average velocity v2 of water leaving the nozzle is:
v
v
a. v1
b. 1
c. 3v1
d. 1
e. 9 v1
9
3
Apply the equation of continuity in the form A1 v1 = A2 v2
Where v1 = v (the average velocity in the pipe) and
v2 (the average velocity in the nozzle) =?
A1 = π (r)2
A2 = π (r/3)2
Thus π (r)2 v = π (r/3)2 v2 = 1/9 π (r)2 v2
v2 = 9 v
Q4.
Two circular metal plates capacitor with radius 0.2 m are separated in vacuum by 10-3 m. The
capacitance of this parallel plates capacitor is: (ε0=8.85×10-12 C2N-1m-2)
b. 1.1x10-9F
c. 8.85x10-11F
e. None of the above
a. 2.8x10-10F
d. 3,55 x10-10F
C= (ε0A
where A= πr2
,
=[(8.85x10-12 C2N-1m-2)3.14x(0.2m)2] 10-3m
=1.1x10-9F
Q5
The number of electrons flowing through a battery that delivers a current of 5.0 A
for 12s is:
a. 4
b. 3.75 × 1020
c. 7.83 × 1015
d. 6.54 × 1013
e. 56
Q6
A room air conditioner uses 500 W of electrical power. If it operates during 20 hours per day
and the energy costs 8 Halala per kilowatt hour, the daily operating cost is:
b. 1 SR
c. 8 SR
d. 0.8 S R
e. 1.6 S R
a) 10 SR
The daily operating cost= (0.5kW)(20 h)(0.08 R/kwh)= 0.8 SR
Q1
Q2
Q3
Q4
Q5
Q6
d
e
e
b
b
d
Q9 -A 4.0-kg block initially at rest is pulled
to the right along a horizontal, frictionless
surface by a constant horizontal force of 9 N.
Find the speed of the block after it has moved
5.0 m.
F
∆x
Solution
The work done by this force is
W=F∆
∆x = (9 N) (5.0 m) = 45 J
( 0.75 point )
Using the work-kinetic energy theorem and noting that the initial kinetic energy is zero, we
obtain:
( 0.75 point )
( 0.5 point )
Q10
In the following distribution
of charges, the electric field
and potential at point P midway
between the two fixed charges
are respectively:
Solution
3.375x103 NC-1
-3x10-6 C
P
+
( 0.75 point )
E= k[
=
+3x10-6
4m
,
( 0.25 point )
Q11
and
are
Two resistors
arranged in a circuit that carries a total current
of 9 A as shown in the figure. What are the
current through
and the voltage across ?
4m
( 0.75 point )
Vp= k[
= 0 volt
( 0.25 point )
I=9 A
R2
Solution
The equivalent resistor R for the two parallel resistors R1=2Ω and R2=4Ω is
(0.75 point)
The voltage across the resistor R1 is the same as that across R and therefore
(0.75 point)
The current through the resistor R2 is
(0.5 point)