Answer:
= 829.69 Watt
≅ 830 Watt
Explanation:
Given that,
Velocity of air flow = 12.5m/s
Rate of flow of air = 9m³/s
Density of air = 1.18kg/m³
power by kinetic energy = 1/2(mv²)
mass = density × volume
m = 1.18 × 9
= 10.62 kg/s
power = 1/2 mV²
= 1/2 (10.62 × 12.5²)
= 829.69 Watt
≅ 830 Watt
Flow rate
u
=
9
m
3
/
s
Velocity of the air
V
=
8
m/s
Density of the air
ρ
=
1.18
kg
/
m
3
Answer:opposite
Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
<h2>Answer:</h2>
<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>
<h3>Explanation:</h3>
Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.
In the given question the circular displacement is 25 rounds around a central point.
The angular displacement is measured in degrees and 1 round is equal to 360 degrees.
25 Rounds = 25 × 360 = 9000 degrees.
Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.
In SI,angular velocity is represented in radians per second.
So, 1 radian = 57.29 degrees
Angular velocity = 15.78 radians per second
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
