Answer:
the length of stretched spring in cm is 22
Explanation:
given information:
spring length, x1 = 20 cm = 0.2 m
force, F = 100 N
the length of spring streches, x2 = 22 cm = 0.22 m
According to Hooke's law
F = - kΔx
k = F/*=(x2-x1)
= 100/(0.22 - 0.20)
= 5000 N/m
if the spring is now suspended from a hook and a 10.2-kg block is attached to the bottom end
m = 10.2 kg
W = m g
= 10.2 x 9.8
= 99.96 N
F = - k Δx
Δx = F / k
= 99.96 / 5000
= 0.02
Δx = x2- x1
x2 = Δx + x1
= 0.20 + 0.02
= 0.22 m
= 22 cm
For some reasons, no. If the driver looks focused and has experience, then it would be okay. Again, it could be dangerous if you bump into the truck, it would cause damage to you and your passengers.
Mostly, I would agree with 'No'. :)
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
Yes it is valid all the times under the consideration of acceleration due to gravity .it is not valid on space where there is no influence of gravity
