100cos30° + 80cos120° + 40cos233° + Dx = 0
<span>Dx = -22.53 N </span>
<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>
<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>
<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
Answer:
C) 1 s
Explanation:
The period of a mass-spring system is given by the formula:

where
m is the mass hanging on the spring
k is the spring constant
As we can see from the equation above, the period of the system does NOT depend on the initial amplitude of the oscillation. Therefore, even if the initial amplitude is changed from 5 cm to 10 cm, the period of the system will remain the same, 1 s.
Answer:
Gravity
because it's factorised by mass of a body.
For other forces, they deal with charges of negligible mass and weights
Fc=mv^2/r so we get
2000kg*(25m/s)^2/(80m)= 15625N of force
hope this helps! Thank You!!
To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.
Our values are given as


From the continuity equations in pipes we have to

Where,
= Cross sectional Area at each section
= Flow Velocity at each section
Then replacing we have,



From Bernoulli equation we have that the change in the pressure is

![7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)](https://tex.z-dn.net/?f=7.3%2A10%5E3%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%281000%29%28%5B%20%5Cfrac%7B%281.25%2A10%5E%7B-2%7D%29%5E2%20%7D%7B0.6%2A10%5E%7B-2%7D%29%5E2%7D%20v_1%20%5D%5E2-v_1%5E2%29)


Therefore the speed of flow in the first tube is 0.9m/s