Download m/s - Egyptian Language School

Egyptian Language School
Science Department
Physics 1st Sec.
Physics Revision sheet for first sec .
1. Match the following physical quantities with its unit:
Physical quantity
1. Displacement.
2. Temperature.
3. Work.
4. Gravitational constant.
5. Density.
6. Force.
7. Acceleration.
8. Mass.
9. Kinetic energy.
Unit
a. Kelvin.
b. Joule.
c. Kg.m/𝑠𝑒𝑐 2
d. N.m2.Kg-2
e. Kg/m3
f. N
g. m/s2
h. meter
i. Kg
j. N.m
2. Compare between:
a. Periodic and translational motion
b-uniform and variable velocity
c. Potential and kinetic energy
3. Complete:
1- ……………………… is equal to 1/86400 of the average solar day.
2-From uses of satellites …………………..and………………………
3-Energy is a …………………… quantity ,while force is ………………………. Quantity.
4- The ratio between potential and kinetic energies at the midpoint of fall equals ……………..
5-the kinetic energy of an object is 4 J. The kinetic energy if its speed is doubled equals
………………..
6-energy is neither created nor destroyed but changes from one form to another is the law of
……………………………………………………………………………..
7-the work done by a force of 1 Newton to move an object through a displacement of one
meter in the direction of force is ………………………..
4-Complete the table
The physical
quantity
Relation to other
quantities
work
Force × displacement
Joule
Mass×velocity
Kg.m/s
Density (ρ)
Mass/volume
Kg/m3
Speed (v)
Distance/ time
Unit
L / T = LT-1
LT-1 /T = LT-2
Acceleration (a)
Force (F)
Dimensional formula
m/s2
Mass x acceleration
N
5- Solve the following problems:
1- A student measured the length of a pencil. It was found 13.9 cm, meanwhile
its actual length is 14cm. Estimate both the absolute error and relative error.
(0.1,0.7%)
2- A car of mass 1 ton starts motion from rest with uniform acceleration of
2 m/s2. Find its kinetic energy: a- Before it starts motion
b- After 10 sec.
3- A body is projected upwards; it reaches the max height after 3 sec. find max height
reached.(g=10m/s2)
4- An object of mass 5Kg falls freely from 10m high above the ground. Find
its kinetic energy when it becomes at 3m high above the ground given that
free fall acc =10 m/s2
5- An object of mass 50Kg has accelerated from rest till its velocity becomes
5m/s after 5 seconds .find the work done.
6- A crane pulls a car with a force 3000N to accelerate it at 3m/s 2 find the
mass and weight of the car.
Good Luck
Model answers of revision sheet
1(1, h) - (2, a) - ( 3 , j) - (4 ,d ) - (5 , e ) – (6 , f and c) - (7 , g) - ( 8 , I ) - (9 , b)
2a-Periodic motion
-Motion repeats itself at equal
intervals of time.
-examples :circular motion and
vibrational motion
Translational motion
-motion which is characterized by
having a starting and end point.
-examples: motion in a straight line
(car or train)
b- Uniform velocity
-The velocity in which the object
moves through equal displacements in
equal intervals of time.
-show by graph
c-Kinetic energy
-Energy gained by an object due to its
motion
-depends on weight and height
3-
Variable velocity
-the velocity in which the object moves
through unequal displacement in equal
intervals of time
-show by graph
Potential energy
Energy stored in an object due to its
position
-depends on mass and velocity
1-the standard second
4- 1:1
5- 16 J
2- Spying and communication
3-scalar - vector
6-Law of conservation of energy 7- joule
4The physical
quantity
Relation to other
quantities
Dimensional formula
Unit
work
Force × displacement
M.L2.T -2
Joule
Momentum
Mass × velocity
M.L.T-1
Kg.m/s
Density (ρ)
Mass/volume
Speed (v)
Distance/ time
Acceleration (a)
Force (F)
Velocity/time
Mass x acceleration
-3
M.L
L / T = LT-1
LT-1 /T = LT-2
M.L/T-2
Kg/m3
m/s
m/s2
N
51- Absolute error = 14-13.9 = 0.1
Relative error =0.1/14 =0.007 =0.7%
2- m= 1ton =1000Kg vi =0
a= 2m/s2
a-before it starts motion v=0 then kinetic energy =0
b- after 10 seconds
vf =vi + at
vf= 0+2×10
v=20m/s
K.E.=1/2mv2 K.E.= 1/2×1000×202
K.E.=200000 joule
g= -10m/s2
3- at max height vf=0
Vf= vi +at
0 = vi + (-10)×3
d=vit +1/2gt2
t=3 s
vi=30 m/s
d=30×3 +1/2 ×(-10)32
4-m=5Kg weight =5×10 =50N
d=45 m
h=10m
At the highest point M.E. = P.E.
P.E.=W× h
P.E. =50×10 =500N M.E.=500N
AT 3m height
M.E.=P.E +K.E.
5- m= 50Kg
P.E. =3×50 =150 J
vi=0
vf=5m/s
t=5 s
d= vit +1/2 a t2
W=F × d
F=m×a
a=vf-vi/t
a=5-0/5 a=1 m/s2
F=m.a F=50× 1=50 N
d=0+0.5×1×52 d=12.5 m
W=F × d =50 ×12.5 = 625 Joule
6-F=3000N a=3m/s2
m=F/a
m=1000Kg
W= m×g =1000×10 = 10000 joule
K.E.= 500-150 =350 J