150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag
1 answer:
Answer:
the friction force on this box is closest to 45.9 N.
Explanation:
given data
Weight of the box W = 150 N
accelerating uniformly = 3.00 m/s²
coefficient of kinetic friction = 0.400
coefficient of static friction = 0.600
solution
we know box does not move relative to the wagon
we get here friction force that is express as
friction force = mass × acceleration ..............1
here mass = weight ÷ g
mass = = 15.3 kg
put value in equation 1 we get
friction force = 15.3 kg × 3
friction force = 45.9 N
So, the friction force on this box is closest to 45.9 N.
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Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
Explanation:
The energy needed to ionize a hydrogen atom in the ground state is:
The energy of the photon is related to the wavelength by
where
h is the Planck constant
c is the speed of light
is the wavelength
Solving the formula for the wavelength, we find