Answer:
31m/s
23.17°
Explanation:
Given the following :
From the diagram attached :
AB = 15m/s, BC = 18m/s, AC = a = resultant
Resolving Velocity into both vertical and horizontal component.
Kindly see attached picture for detailed explanation.
Answer:
v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²
Explanation:
This is a missile throwing exercise.
On the x axis there is no acceleration so the velocity on the x axis is constant
v₀ₓ = 10 m / s
On the y-axis velocity is affected by the acceleration of gravity, let's use the equation
v_y =
- g t
at the highest point of the trajectory the vertical speed must be zero
v_y = 0
therefore the velocity of the body is
v = (10 i ^ + 0j ^) m / s
the acceleration is
a = (0 i ^ - g j⁾
a = (0i ^ - 9.8 j ^) m / s²
The correct answer is D. the star is getting farther away.
When the light has red-shifted, it shows that everything is moving away from a certain point. This is known as the Doppler effect and proves the Big Bang (which would be the center point that everything is moving away from). Just as a 'for your information', blue shift is when the star is moving towards us.
Hope I helped :)
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where



D it causes more useful traits to be passed to offspring