It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?
Answer:
Target ceiling. the upper limit of your physical activity. Target fitness zone. Above the threshold of training and below the target ceiling.
Hope this helps. Can u give me brainliest
Explanation:
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
Answer:
1) v = 0.45 m/s
2) v = 0.65 m/s
3) v = 0.75 m/s
Explanation:
1) We can find the speed of the object by conservation of energy:
Where:
k: is the spring constant = 280 N/m
v: is the speed of the object =?
m: is the mass of the object = 5.00 kg
x: is the displacement of the spring
2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:
3) When the object is at the equilibrium position, the speed of the object is:
I hope it helps you!
