Answer:
The distance between knothole and the paint ball is 0.483 m.
Explanation:
Given that,
Height = 4.0 m
Distance = 15 m
Speed = 50 m/s
The angle at which the forester aims his gun are,
Using the equation of motion of the trajectory
The horizontal displacement of the paint ball is
Using the equation of motion of the trajectory
The vertical displacement of the paint ball is
Put the value into the formula
We need to calculate the distance between knothole and the paint ball
Hence, The distance between knothole and the paint ball is 0.483 m.
Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:
We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
![|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}](https://tex.z-dn.net/?f=%7Cv_%7Bhawk%7D%7C%3D%5Csqrt%5B%5D%7B6%5E2%2B2%5E2%7D%3D%5Csqrt%7B40%7D)
≈
And the angle with respect to the east should be with:
2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one
