Answer:
so angular velocity is 7.13128 sec−1
Explanation:
velocity v = 2.2 m/s
displacement s = 220 mm = 0.220 m
distance d = 510 mm = 0.510 m
to find out
angular velocity
solution
we know that
angular velocity will be velocity ( v) / (displacement² + distance²) .....1
now put all these value in equation 1 and we get angular velocity i.e.
angular velocity = velocity ( v) / (displacement² + distance²)
angular velocity = 2.2 / (0.22² + 0.51²)
angular velocity = 2.2 / 0.3085
angular velocity = 7.13128
so angular velocity is 7.13128 sec−1
I think it’s B I could be wrong but I tried lol
If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Answer:
c)
V_local = -x/t^2
V_convec = x/t^2
d)
a = V_local + V_convec = 0
e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases
Explanation:
Hi!
You can see plots for a) and b) attached on this document
c)
The local acceleration is just teh aprtial derivative of the velocity with respect to t:
And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:
d)
Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign
e)
This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases, the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
