Answer:
TECHNICIAN B
Explanation: Drive shaft is an elongated part of the drive system made of steel of a vehicle that transmits the <em>ENGINE TORQUE to the WHEELS</em> of a vehicle. For proper inspection a technician is expected to completely remove the driveshaft to check all its compactment. It is Cylindrical and usually used for <em>REAR-WHEEL-VEHICLES</em>. Drive shafts are known by different names like propeller shaft,Cardan shaft etc it has different types which range from
<em>One-piece drive shaft
,Two-piece drive shaft to Slip-in-tube drive shaft
.</em>
Answer:
u = 104.68 m/s
Explanation:
given,
horizontal distance = 150 m
elevation of 12.4 m
angle = 8.6°
horizontal motion = x = u cos θ. t .............(1)
vertical motion =
................(2)
from equation(1) and (2)
..........{3}
u = 104.68 m/s
The initial speed of the ball is u = 104.68 m/s
<h3><u>Answer;</u></h3>
D) Standing wave
<h3><u>Explanation;</u></h3>
- Standing wave also called stationary wave is a wave which oscillates in time but whose peak amplitude profile does not move in space.
- A standing wave pattern is a vibrational pattern created within a medium when the vibrational frequency of the source causes reflected waves from one end of the medium to interfere with incident waves from the source.
- Examples of standing waves include the vibration of a violin string and electron orbitals in an atom.
Density: g/mL, kg/cubic meter
Volume: L, teaspoon
Mass: g, MeV/sq. C
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
