Answer:
A. 112 J
Explanation:
KE = ½mv² = ½(0.14)40² = 112 J
Answer:
<em>11.06m/s²</em>
Explanation:
According to Newtons second law of motion

Given
Mass m = 17kg
Fm = 208N
theta = 36 degrees
g = 9.8m/s²
a is the acceleration
Substitute
208 - 0.148(17)(9.8)cos 36 = 17a
208 - 24.6568cos36 = 17a
208 - 19.9478 = 17a
188.05 = 17a
a = 188.05/17
a = 11.06m/s²
<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>
Answer:
1. 610,000 lb ft
2. 490 J
Explanation:
1. First, convert mi/hr to ft/s:
100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s
Now find the kinetic energy:
KE = ½ mv²
KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²
KE = 610,000 lb ft
2. KE = ½ mv²
KE = ½ (5 kg) (14 m/s)²
KE = 490 J
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ
............1
put here value
= 1.75×
× 
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 ×
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 ×
v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 ×
v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds
Answer:
he spring provides the controlling torque. The air friction induces the damping torque, which opposes the movement of the coil. The repulsion type instrument is a non-polarized instrument, i.e., free from the direction of current passes through it. Thus, it is used for both AC and DC