The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:
ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)
or equivalently,
(see the attached graphic)
We have
- ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or
- wind (relative to Earth) = 5.0 m/s due East, or
- ducks (relative to earth) = some speed <em>v</em> due South, or
Then by setting components equal, we have
We only care about the direction for this question, which we get from the first equation:
or approximately 136º or 224º.
Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want
which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).
So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.
Answer:
Balancing of forces,
In the X-direction:
-Tcos4.5^o +Tcos4.5^o=0
In the Y-direction:
Tsin4.5^o +Tsin4.5^o-m*g=0
2Tsin4.5^o=15*9.81
T=(15*9.81)/(2sin4.5^o)
=937.75 N
Therefore, tension in the rope is 937.75 N.
Answer:
V = -RC (dV/dt)
Solving the differential equation,
V(t) = V₀ e⁻ᵏᵗ
where k = RC
Explanation:
V(t) = I(t) × R
The Current through the capacitor is given as the time rate of change of charge on the capacitor.
I(t) = -dQ/dt
But, the charge on a capacitor is given as
Q = CV
(dQ/dt) = (d/dt) (CV)
Since C is constant,
(dQ/dt) = (CdV/dt)
V(t) = I(t) × R
V(t) = -(CdV/dt) × R
V = -RC (dV/dt)
(dV/dt) = -(RC/V)
(dV/V) = -RC dt
∫ (dV/V) = ∫ -RC dt
Let k = RC
∫ (dV/V) = ∫ -k dt
Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.
In V - In V₀ = -kt
In(V/V₀) = - kt
(V/V₀) = e⁻ᵏᵗ
V = V₀ e⁻ᵏᵗ
V(t) = V₀ e⁻ᵏᵗ
Hope this Helps!!!
Answer:
Answer in Explanation
Explanation:
Whenever we talk about the gravitational potential energy, it means the energy stored in a body due to its position in the gravitational field. Now, we know that in the gravitational field the work is only done when the body moves vertically. If the body moves horizontally on the same surface in the Earth's Gravitational Field, then the work done on the body is considered to be zero. Hence, the work done or the energy stored in the object while in the gravitational field is only possible if it moves vertically. This vertical distance is referred to as height. <u>This is the main reason why we require height in the P.E formula and calculations.</u>
The derivation of this formula is as follows:
Work = Force * Displacement
For gravitational potential energy:
Work = P.E
Force = Weight = mg
Displacement = Vertical Displacement = Height = h
Therefore,
P.E = mgh
Answer:
Explanation:
Remark
This is a momentum question. Both cars are sitting still (v1 and v2 are both 0) to start with). When the spring is sprung), they both move but in opposite directions.
Let us say that v4 is minus.
Equation
0 = m3*v3 - m4* v4
Givens
m3 = 1750 kg
m4 = 1000 kg
v3 = 4 m/s
Solution
m3*4m/s - m4*x = 0
1750 * 4 - 1000*x = 0
1750 * 4 = 1000x
7000 = 1000 x
7000/1000 = x
x = 7 m/s
