Answer:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Answer:
Fr = 26.83 [N]
Explanation:
To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.
![F_{r}=\sqrt{(12)^{2}+(24)^{2} } \\F_{r}=26.83[N]](https://tex.z-dn.net/?f=F_%7Br%7D%3D%5Csqrt%7B%2812%29%5E%7B2%7D%2B%2824%29%5E%7B2%7D%20%20%7D%20%5C%5CF_%7Br%7D%3D26.83%5BN%5D)
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is 
The focal length of A is 
The theoretical angular magnification is mathematically represented as


Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range
If<span> a neutral </span>object loses<span> some </span>electrons<span>, </span>then<span> it will possess more protons</span>