Answer:
a) 
, b) 
, c) 
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:
b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:
c) The spring constant of the gun is:
To determine the centroid of the object first moment of area is used.
To predict the resistance of a shape to bending and deflection which are directly proportional, second moment of area is used.
Answer:
S = V t where S is the horizontal distance traveled
1/2 g t^2 = H where H is the vertical distance traveled
t^2 = 2 H / g
V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations
tan theta = H / S
V^2 = S g / (2 tan theta)
Using S = L cos theta
V^2 = L g cos theta / (2 tan theta)
Giving V in terms of L and theta
We know, F = m * a
F = 10 * 5
F = 50 N
In short, Your Answer would be 50 Newtons
Hope this helps!
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
