To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
No, because sometimes you have to stop at stop signs and stop lights.
Simply draw the vector with the given coordinates.
Answer: 18.27°
Explanation:
Given
Index of refraction of blue light, n(b) = 1.64
Wavelength of blue light, λ(b) = 440 nm
Index of refraction of red light, n(r) = 1.595
Wavelength of red light, λ(r) = 670 nm
Angle of incident, θ = 30°
Angle of refraction of red light is
θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1
So that,
θ(r) = sin^-1 [(1 * sin 30) / 1.595]
θ(r) = sin^-1 (0.5 / 1.595)
θ(r) = sin^-1 0.3135
θ(r) = 18.27°
Answer:
7m/s^2
Explanation:
using v=u+at
since the car started from rest, u=0 , v=14m/s t=2s
a =acceleration.
14=0+a×2
14=0+2a
14=2a
a= 14/2 =7
a=7m/s^2
