Straight line distance between their home and the store can be solve using cosine law. first is solve the angle which is180 - 45 = 135 degree
C^2 = a^2 + b^2 - 2ab cos(135)c^2 = 5^2 + 5^2 - 2(5)(5) cos(135)c^2 = 85.35c = 9.24 km is the straight line distance between their home and the store
Answer:
Velocity of Object with 2 kg= 3.390 m/s
Velocity of Object with 3 kg= 3.404 m/s
Explanation:
From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.
We also know that momentum of a closed system is conserved.
Initial momentum along the x-axis = 2*5.5 = 11
Initial momentum along y-axis = 0
Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).
Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)
Solve both the equations simultaneously for a and b.
As wavelength increase, frequency decrease
The universe is made up of baryonic matter which is a combination of protons, electrons, and neutrons, and also dark matter and dark energy.
Galaxies are comprised of stars, while the universe is comprised of galaxies.
I hope my answer has come to your help. God bless and have a nice day ahead!
<h2>
Answer: (a)t=0.553s, (b)x=110.656m</h2>
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:
x-component:
(1)
Where:
is the bullet's initial speed
because we are told the bullet is shot horizontally
is the time since the bullet is shot until it hits the ground
y-component:
(2)
Where:
is the initial height of the bullet
is the final height of the bullet (when it finally hits the ground)
is the acceleration due gravity
<h2>Part (a):</h2>
Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:
(3)
(4)
Finding :
(5)
Then we have the time elapsed before the bullet hits the ground:
(6)
<h2>Part (b):</h2>
For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:
(1)
Substituting the knonw values and the value of found in (6):
(7)
(8)
Finally:
