The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;

- Mass of second block,

- Tension on first rope;

- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;

Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
Answer:
a. The station is rotating at 
b. the rotation needed is 
Explanation:
We know that the centripetal acceleration is

where
is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).
<h3>a. </h3>
The rotational speed is :




Knowing that there are
in a revolution and 60 seconds in a minute.


<h3>b. </h3>
The rotational speed needed is :




Knowing that there are
in a revolution and 60 seconds in a minute.


The correct answer is the third, It reflects the green light waves and absorbs most of the rest.
Answer:
The power
Explanation:
We know that the work definition is given by the following expression:
W = F * d
where:
F = force [Newtons] [N]
d = distance [meters] [m]
W = work [Joules]
And the expression that defines the work done by unit of time is called - <u>Power</u>, therefore:
P = W/t
where:
P = power [watts] [w]
W = work [Joules] [J]
t = time [seconds] [s]
Answer:
1.7 m/s²
Explanation:
d = length of the ramp = 13.5 m
v₀ = initial speed of the skateboarder = 0 m/s
v = final speed of the skateboarder = 7.37 m/s
a = acceleration
Using the equation
v² = v₀² + 2 a d
7.37² = 0² + 2 a (13.5)
a = 2.01 m/s²
θ = angle of the incline relative to ground = 29.9
a' = Component of acceleration parallel to the ground
Component of acceleration parallel to the ground is given as
a' = a Cosθ
a' = 2.01 Cos29.9
a' = 1.7 m/s²