Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
![K = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg
Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here
Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)
Answer:
Resultant Force=33.8 lb
Angle=67.2°
Explanation:
Given data
Fa=22 lb
Fb=16 lb
Θ=55⁰
To find
(i) Resultant Force F
(ii)Angle α
Solution
First we need to represent the forces in vector form
![\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20F_%7B1%7D%3D22j%5C%5C%20F_%7B2%7D%3Du%2Bv%5C%5CF_%7B2%7D%3D16sin%2855%29i%2B16cos%2855%29j%5C%5CF_%7B2%7D%3D16%280.82%29i%2B16%280.5735%29j%5C%5CF_%7B2%7D%3D13.12i%2B9.176j)
Total Force
![F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j](https://tex.z-dn.net/?f=F%3DF_%7B1%7D%2BF_%7B2%7D%5C%5C%20F_%7B2%7D%3D22j%2B13.12i%2B9.176j%5C%5CF_%7B2%7D%3D13.12i%2B31.176j)
The Resultant Force is given as
![|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb](https://tex.z-dn.net/?f=%7CF%7C%3D%5Csqrt%7Bx%5E%7B2%7D%20%2By%5E%7B2%7D%20%7D%5C%5C%7CF%7C%3D%5Csqrt%7B%2813.12%29%5E%7B2%7D%20%2B%2831.176%29%5E%7B2%7D%20%7D%5C%5C%20%7CF%7C%3D33.8lb)
For(ii) angle
We can find the angle bu using tanα=y/x
So
![tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}](https://tex.z-dn.net/?f=tan%5Calpha%20%3D%5Cfrac%7B31.176%7D%7B13.12%7D%5C%5C%20%5Calpha%20%3Dtan%5E%7B-1%7D%20%28%5Cfrac%7B31.176%7D%7B13.12%7D%29%5C%5C%5Calpha%20%3D67.2%5E%7Bo%7D)
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
"A pendulum swinging back and forth" is an example of harmonic motion
X = Xo cos ω t
Explains the back and forth motion of the pendulum