Answer:
part (a)
towards north east direction.
part (b) s = 46.60 m
Explanation:
Given,
- velocity of the river due to east =

- velocity of the boat due to the north =

part (a)
River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,
Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 
Let 'v' be the velocity of the boat relative to the shore.

Let
be the angle of the velocity of the boat relative to the shore with the horizontal axis.
Direction of the velocity of the boat relative to the shore.
part (b)
- Width of the shore = w = 300m
total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken
Let 't' be the total time taken by the boat to cross the width of the river.
Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river
Galaxies are sprawling systems of dust, gas, dark matter, and anywhere from a million to a trillion stars that are held together by gravity. Nearly all large galaxies are thought to also contain supermassive black holes at their centers.
Answer:
V₀ = 5.47 m/s
Explanation:
The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 3.04 m
θ = Launch Angle = 41.7°
V₀ = Minimum Launch Speed = ?
g = 9.81 m/s²
Therefore,
3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)
V₀² = 3.04 m/(0.10126 s²/m)
V₀ = √30.02 m²/s²
<u>V₀ = 5.47 m/s</u>
Answer:
It comes out the positive side of the battery and goes in to the negative side of the battery
Explanation:
There are already electrons in wires in a circuit before you add the battery. By adding the battery, you're giving the electrons the energy it needs to move along the circuit.
In a series circuit, the circuit is one continuous loop so there is only one path for the electrons to go - out of the positive side of the battery and around the circuit then goes back into the negative side of the battery.
However, with a parallel circuit, there are two or more ways the electrons can go so they take the path of least resistance. The electrons still go out the positive side of a battery but along the circuit, the electrons will go through the path of least resistance ( I tend to think of it like a net with holes in it - the lower the resistance the bigger the holes for the electrons to go through so more can fit in a set amount of time ) but the electrons still go out of the positive side and in through the negative
Answer:
The emission spectrum is always the same and can be used to identify the element and part of the Bohr model proposed that electrons in the hydrogen are located in particular orbits around the nucleos are True.
Explanation:
The Niels Bohr and quantic mecanic theorys are both based on the study of atomics spectrums. The atomic spectrum is a characteristic pattern of a light wavelenght emited wich is unique to each element.
<u>For example</u>, if we put some low pressure hydrogen in a glass tube and in the tp of the glass we apply a voltage big enough to produce a electric current in the hydrogen gas, the tube its going to emit light wich have a color dependig of the gas element in the interior. If we observe this light with a spectrometer we are going to see shining lines and each one of this lines have a wavelenght and diferent colors. This lines are called emission spectrum and the wavelength of that spectrum are unique to eache element.
<u>Summering up, </u>we can identify elements using the emission spectrum because any element produces the same spectrum than other element.
According to Niels Bhor theory the electron only can be in especific discret ratios to the nucleus. Where this electron moves himself in circukar orbits under the influence of the Coulomb attraction force.