A vice pushes on a system of three boards, each oriented vertically and held up by horizontal forces. The outer boards weigh 90
N and the inner board weighs 118 N. The coefficient of friction between the inner and the outer boards is 0.67. What is the magnitude of one of the compression forces acting on either side of the inner board?
1 answer:
Answer:
88.1 N
Explanation:
As shown in the free body diagram attached to the question, the only forces acting on the inner block include in the required vertical direction.
- Its Weight.
- Frictional forces as a result of the Two compression forces.
If the block is not to slip off, the weights must match the two frictional forces
Let the compressive forces be F and frictional force be Fr
Fr = μ F
Sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
Each of the compression forces is 88.1 N
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Answer:
a) = 1.9 eV
= 3.04 10⁻¹⁹ J,b ) This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.
Explanation:
a) To calculate the maximum kinetic energy of the expelled electrons let's use the relationships of the photoelectric effect
= h f - Φ
Where K is the kinetic energy, h the Planck constant that is worth 6.63 10⁻³⁴ Js, f the frequency and Φ the work function
The speed of light is related to wavelength and frequency
c = λ f
Let's analyze the work function, it is the energy needed to start an electron from a metal, in this case to start an electron from a hydrogen atom its fundamental energy is needed, so
Φ= E₀ = 13.6 eV
let's replace and calculate the energy of the incident photon
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/80 10⁻⁹
E = 2,486 10⁻¹⁸ J
Let's reduce to eV
E = 2,486 10⁻¹⁸ (1 eV / 1.6 10⁻¹⁹)
E = 15.5 eV
Now we can calculate the kinetic energy
= h c / f - fi
= 15.5 -13.6
= 1.9 eV
b) Extra energy = 10.2 eV
The total kinetic energy of electrons is
Total kinetic energy = 1.9 +10.2 = 12.1 eV
For the calculation we are assuming that all the electors are in the hydrogen base state, but for temperatures greater than 0K some electors may be in some excited state, so less energy is needed to tear them out of hydrogen atom.
Let's analyze this possibility
ΔE = E photon - Total kinetic energy electron
ΔE = 15.5 - 12.1
ΔE = 3.4 eV
If we use the Bohr ratio for the hydrogen atom
= 13.606 / n2
n = √ 13.606 / En
n = √ (13606 / 3.4)
n = 2
This means that some electrons are at the first excited level of the hydrogen atom, which is highly likely as the temperature rises.
Answer:
0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that
Where
u(0)=0
Substitute the value
Substitute u'(0)=10
Substitute the values
Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec