Practice Test 3
Choose the best answer
- Convert 3.2 radians to degrees.
- 344.402 deg
- 264.523 deg
- 63.957 deg
- 204.059 deg
- 183.346 deg
Answer: E
- An object is revolving in a circular path of radius 2 m
with a uniform speed of 21 m ⁄ s. How many
cycles does it execute per second.
- 1.671 Hz
- 1.271 Hz
- 0.354 Hz
- 2.507 Hz
- 2.75 Hz
Answer: A
- An object is revolving in a circular path of
radius 6 m with a uniform speed. If it makes
5 revolutions in 21 seconds, calculate
its speed.
- 1.401 m ⁄ s
- 3.306 m ⁄ s
- 8.976 m ⁄ s
- 15.162 m ⁄ s
- 6.737 m ⁄ s
Answer: C
- An object is revolving in a circular path of
radius 6 m with a uniform speed. If it makes
19 cycles per second, calculate its
centripetal acceleration.
- 72837.281 m ⁄ s 2
- 21929.293 m ⁄ s 2
- 141085.242 m ⁄ s 2
- 85510.253 m ⁄ s 2
- 119937.125 m ⁄ s 2
Answer: D
- Centrpetal acceleration of a circular motion
- is acceleration due to change in angular speed
- is acceleration due to the change in direction
of the velocity.
- is the same with angular acceleration
- is acceleration due to change of speed and
change in the direction of the velocity
- is acceleration due to change of speed
Answer: B
- An object is revolving in a circular path of radius 4 m
with a uniform speed of 16 m ⁄ s. Calculate its
angular velocity.
- 5.487 rad ⁄ s
- 4 rad ⁄ s
- 2.732 rad ⁄ s
- 6.079 rad ⁄ s
- 1.905 rad ⁄ s
Answer: B
- An object is revolving in a circular path of
radius 4.2 m with an angular acceleration of 6 rad ⁄ s 2.
At a time when its angular speed is 15 rad ⁄ s, calculate the net
acceleration of the object.
- 1736.353 m ⁄ s 2
- 519.137 m ⁄ s 2
- 945.336 m ⁄ s 2
- 342.173 m ⁄ s 2
- 759.562 m ⁄ s 2
Answer: C
- Gravitational force between two objects is
- an attractive force which is proportional to the
product of the masses of the objects and
inversely proportional to the square of the
distance separating the two objects.
- a repulsive force which is proportional to the
product of the masses of the objects and
inversely proportional to the distance
separating the two objects.
- an attractive force which is proportional to the
product of the masses of the objects and
inversely proportional to the distance
separating the two objects.
- an attractive force which is proportional to the
sum of the masses of the objects and
inversely proportional to the distance
separating the two objects.
- an attractive force which is proportional to the
sum of the masses of the objects and
inversely proportional to the square of the
distance separating the two objects.
Answer: A
- A satelite of mass 700000 kg is revolving around
earth at an altitude of 5000 km above the surface of
earth. Earth has a mass of 5.98e24 kg and a radius of 6.38e6 m.
Calculate the acceleration with which the satelite is revolving
around earth.
- 3.08 m ⁄ s 2
- 5.178 m ⁄ s 2
- 5.816 m ⁄ s 2
- 0.807 m ⁄ s 2
- 2.023 m ⁄ s 2
Answer: A
- An object of mass 2 kg is revolving in a vertical
circle of radius 1.7 m. Calculate the minimum
speed at the top of the circle by which the
object can make it without the string slacking.
- 4.082 m ⁄ s
- 7.66 m ⁄ s
- 6.595 m ⁄ s
- 3.205 m ⁄ s
- 1.348 m ⁄ s
Answer: A
- The principle of conservation of angular momentum states that
- the angular momentum of an object will be
conserved if the net torque acting on the
object is a constant.
- the angular momentum of an object will be
conserved if all the forces acting on the
object are conservative.
- the angular momentum of an object will be
conserved if the net torque acting on the
object is zero.
- the angular momentum of an object will be
conserved if the net force acting on the
object is zero.
- the angular momentum of an object will be
conserved if the net force acting on the
object is a constant.
Answer: C
- A horizontal lever of length 1.6 m is pivoted
at its mid-point. An downward force of 13 N is
acting at the right end of the lever. Calculate
the torque acting on the lever.
- -15.204 N m
- -5.636 N m
- -17.805 N m
- -4.175 N m
- -10.4 N m
Answer: E
- A uniform horizontal lever of length 3.6 m is
pivoted at its mid-point. The following forces
are acting on the lever: a 0.53 N vertically upward force
acting at the right end of the lever, a 11 N
vertically upward force acting at the left end of the
lever, and a 4 N vertically downward force (with downward
vertical component) that makes an angle of 10 with
the horizontal-right acting at a point on the lever 0.6 m
away from the right end. Calculate the net
torque acting on the lever.
- -2.599 N m
- -22.652 N m
- -5.838 N m
- -19.68 N m
- -16.588 N m
Answer: D
- A uniform horizontal lever of length 5.23 m is pivoted
at its mid-point. A vertically downward force of 6.71 N is
acting at the right end of the lever. An unknown vertically
downward force is acting at a dstance of 0.53 m
to the left of the pivot. If the lever is in equilibrium,
calculate the force exerted by the pivot
(fulcurum) on the lever.
- 39.817 N
- 50.692 N
- 64.449 N
- 55.234 N
- 12.988 N
Answer: A
- The center of gravity of two particles of masses
20.33 kg and 9.87 kg is located at the point ( 3, 3 ) m.
The 20.33 kg particle is located at the point ( -3, -4 ) m.
Find the location of the 9.87 kg particle.
- ( 15.359, 17.418 ) m
- ( 4.899, 8.115 ) m
- ( 11.276, 17.418 ) m
- ( 11.276, 24.661 ) m
- ( 15.359, 24.661 ) m
Answer: A
- An object is said to be in translational equilibrium if
- it is eithe at rest or rotating with a constant
angular acceleration.
- it is either at rest or rotating with a constant
angular velocity.
- it is either at rest or moving with a constant speed
- it is either at rest or moving with a constant acceleration
- it is either at rest or moving in a straight
line with a constant speed.
Answer: E
- An object of mass 4.2 kg is revolving in a circular
path of radius 2.4 m with an angular acceleration of
10.6 rad ⁄ s 2. Calculate the tangental force acting on
the object.
- 119.078 N
- 174.694 N
- 153.548 N
- 61.169 N
- 106.848 N
Answer: E
- Three particles of masses 10.3 kg, 1.5 kg and 5.7 kg are located
at the points ( 9, 5 ) m, ( 13, 2 ) m, and ( -10, -4 ) m respectively.
Calculate the angular momentum of this system of
particles if they are revolving around the y-axis with
an angular velocity of 21 rad ⁄ s.
- 49547.702 J s
- 34813.8 J s
- 25136.377 J s
- 4815.419 J s
- 53622.253 J s
Answer: B
- A spherical object of mass 4 kg and radius
0.028 m is rolling down an inclined plane of length
2.5 m that makes an angle of 80 deg with the ground.
Calculate its speed by the time it reaches the ground
( Ispehre = 2MR 2 / 5 )
- 8.542 m ⁄ s
- 5.871 m ⁄ s
- 3.746 m ⁄ s
- 9.278 m ⁄ s
- 7.042 m ⁄ s
Answer: B
- The angular momentum of a spherical object revolving
about an axis passing through its center increased
from 7 J s to 29 J s in 3 s. Calculate the
torque acting on it.
- 2.099 N m
- 4.267 N m
- 12.01 N m
- 7.333 N m
- 10.738 N m
Answer: D
- A skater whose moment of inertia is 7.25 kg m 2 is revolving
with an angular speed of 5.9 rad ⁄ s. She decreases her
angular speed to 4.6 rad ⁄ s by extending her hands.
Calculate her new moment of inertia.
- 12.659 kg m 2
- 9.299 kg m 2
- 1.318 kg m 2
- 6.883 kg m 2
- 14.483 kg m 2
Answer: B
- The kind of stress where the force is applied
perpendicularly on the entire surface area
of an object is called
- parallel stress
- tensile stress
- shear stress
- normal stress
- bulk stress
Answer: E
- An aluninum wire has a length of 8 m and a
cross-sectional radius of 0.0025 m. Calculate the change in
its length when an object of weight 60 N hangs from the wire.
(modulus = 7e10 Pa)
- 47.722e-5 m
- 24.779e-5 m
- 40.209e-5 m
- 34.923e-5 m
- 51.296e-5 m
Answer: D
- At what depth in an ocean would the pressure be
5 times atmospheric pressure at sea level. (Assume the
density of the ocean to be 1000 kg ⁄ m 3 )
- 29.471 m
- 41.347 m
- 76.998 m
- 57.674 m
- 45.626 m
Answer: B
- In an open manometer filled with mercury (densty =
13600 kg ⁄ m 3 ), the level of mercury column in the air side is
0.09 m higher than that in the gas side. Determine the pressure
of the gas. Atmospheric pressure is 1e5 Pa.
- 14394.24 Pa
- 88004.8 Pa
- 13194.72 Pa
- 11995.2 Pa
- 111995.2 Pa
Answer: E
- Determine the atmospheric pressure at a place
where a mercury (density = 13600 kg ⁄ m 3 ) barometer
rises by 0.4 m.
- 53312 Pa
- 6013.585 Pa
- 81281.95 Pa
- 45180.263 Pa
- 63628.154 Pa
Answer: A
- Two tubes of different cros-sectional areasare
connected together horizontally. Which of
the following is a true statement about a
fluid flowing through the tubes.
- The speed of the fluid in the narrower tube is
greater than the speed of the fluid in the
wider tube.
- The pressure of the fluid in the narrower tube
is greater than that in the wider tube.
- The amount of fluid that leaves the wider tube
is greater than the amount of fluid that
enters the narrower tube in the same interval
of time.
- The speed of the fluid in the narrower tube is
less than the speed of the fluid in the wider tube.
- The amount of fluid that leaves the wider tube
is less than the amount of fluid that enters
the narrower tube in the same interval of time.
Answer: A
- An object of volume 9e-6 m 3 and density 9000 kg ⁄ m 3 is
immersed inside a fluid of density 2500 kg ⁄ m 3. Calculate
the force exerted by the fluid on the object.
- 0.31 N
- 0.355 N
- 0.168 N
- 0.221 N
- 0.038 N
Answer: D
- An object of volume 7e-5 m 3 and density 750 kg ⁄ m 3 is floating in a
fluid of density 2500 kg m 3. Calculate the volume of the object
exposed above the surface of the fluid.
- 0.543e-5 m 3
- 4.9e-5 m 3
- 2.797e-5 m 3
- 7.178e-5 m 3
- 9.16e-5 m 3
Answer: B
- Tube 1 and tube 2 are connected together. A fluid
flowing through these tubes has a speed of 4 m ⁄ s in tube 1
and a speed of 19 m ⁄ s in tube 2. Calculate the ratio
of the cross-sectional radius of tube 2 to the cross-sectional
radius of tube 1.
- 0.232
- 0.687
- 0.459
- 0.086
- 0.534
Answer: C
- Two tubes of different cross-sectional radii are
connected together with the first tube
elevated 0.7 m above the second tube. A fluid of density
2000 kg ⁄ m 3 enters the first tube with a speed of 0.75 m ⁄ s. The
cross-sectional radius of the first tube is 0.02 m
and that of the second tube is 0.16 m. If the pressure
of the fluid in the first tube is 2e3 Pa, calculate
the pressure of the fluid in the second tube.
- 4040.917 Pa
- 24330.974 Pa
- 20620.682 Pa
- 16282.363 Pa
- 10684.351 Pa
Answer: D