If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and 
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Answer:
1.95m/s
Explanation:
Please view the attached file for the detailed solution.
The following were the conversion factors used in order to express all quatities in SI units:
It is 2.) Cut the DNA into fragments
Answer:
10
Explanation:
Lets say that a,b,c,d,e are the five allowed energy states in order of decreasing energy. Then the number of possible different spectral lines comes from the electron dropping from a high state to a lower state. So, they will do so in following ways:
a - b
a - c
a - d
a- e
b - c
b - d
b- e
c - d
c- e
d- e
Ans- Ten different possible energy jumps giving six different colors or lines in the spectrum.
The sound wave will have traveled 2565 m farther in water than in air.
Answer:
Explanation:
It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.
Distance = Velocity × Time.
So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.
As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.
Distance = 343 × 2.25 =771.75 m
And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.
Distance = 1483×2.25=3337 m.
Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.
Difference in distance covered in water and air = 3337-772 m = 2565 m
So the sound wave will have traveled 2565 m farther in water than in air.
