Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
|
|
B1 at 20km/h
|
|
V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
Answer:
Explanation:
1 )
Here
wave length used that is λ = 580 nm
=580 x 10⁻⁹
distance between slit d = .46 mm
= .46 x 10⁻³
Angular position of first order interference maxima
= λ / d radian
= 580 x 10⁻⁹ / .46 x 10⁻³
= 0.126 x 10⁻² radian
2 )
Angular position of second order interference maxima
2 x 0.126 x 10⁻² radian
= 0.252 x 10⁻² radian
3 )
For intensity distribution the formula is
I = I₀ cos²δ/2 ( δ is phase difference of two lights.
For angular position of θ1
δ = .126 x 10⁻² radian
I = I₀ cos².126x 10⁻²/2
= I₀ X .998
For angular position of θ2
I = I₀ cos².126x2x 10⁻²/2
= I₀ cos².126x 10⁻²
Static Friction
It is the friction that exists between a stationary object and the surface on which it's resting.
Sliding friction
It is the resistance created by two objects sliding against each other.
Rolling friction:-
It is the force resisting the motion when a body rolls on a surface.
hope this helps x
The tangent looks good.
The curve is a bit crooked, at the 0.9 and 1.
But overall, cool graph.
Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
