Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule
We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
Here is a similar question brainly.com/question/13338067?referrer=searchResults
Answer:
2.5 kg.m/s
Explanation:
Taking left side as positive while right side direction as negative then
Momentum, p= mv where m is the mass of the object and v is the velocity of travel
Momentum for ball moving towards right side=mv=2.5*-3=-7.5 kg.m/s
Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s
Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s
The period of the wave is the reciprocal of its frequency.
1 / (5 per second) = 0.2 second .
The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.
Wave speed = (frequency) x (wavelength)
= (5 per second) x (1cm) = 5 cm per second
I think it's The fossil record. The same animal fossil is in Africa and South America. The animal could have not swim across so its the fossil record
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
