Answer: let north be+ and south be-
therefore, -1 km is the resultant displacement of the bird
hope it helped u,
pls put thanks and pls mark as the brainliest
^_^
Answer:
mass = 0.18 [kg]
Explanation:
This is a classic problem where we can apply the definition of density which is equal to mass over volume.
![density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]](https://tex.z-dn.net/?f=density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C%5C%5Cwhere%3A%5C%5Cvolume%20%3D%201%20%5Bm%5E3%5D%5C%5Cdensity%20%3D%200.18%5Bkg%2Fm%5E3%5D)
mass = 0.18*1
mass = 0.18 [kg]
Answer:
It will take Andy 1.198minutes to mow the lawn and it will take Brian 1,209.98minutes to mow the same lawn.
Step-by-step explanation:
Using the simultaneous equation concept, let A denote Andy and B be Brian
Andy and Brian can mow the lawn for 1212minutes i.e A+B = 1212..eqn 1
If Brian would mow the lawn by himself in 1010 minutes more than it would take Andy, this means B=1010A...eqn 2.
Substituting eqn 2 into eqn 1
Equation 1 becomes
A+1010A=1212
1011A=1212
A=1212/1011
A=1.198
B = 1010×1.198
B=1,209.98
Therefore, It will take Andy 1.198minutes to mow the lawn and it will take Brian 1,209.98minutes to mow the same lawn.
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ
