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:
32.1 N Please Give Brainliest
Explanation:
force = mass x acceleration
Answer:
t = 5 s
Explanation:
Data:
- Initial Velocity (Vo) = 7 m/s
- Acceleration (a) = 3 m/s²
- Final Velocity (Vf) = 22 m/s
- Time (t) = ?
Use formula:
Replace:
Solve the subtraction of the numerator:
It divides:
How much time did it take the car to reach this final velocity?
It took a time of <u>5 seconds.</u>
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
