A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.7 kg and the sign has a mass of ms = 17.
2 kg. The length of the beam is L = 2.78 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = 33.9°.
What is the tension in the wire? What is the net force the hinge exerts on the beam? The maximum tension the wire can have without breaking is
T=936 N. What is the maximum mass sign that can be hung from the beam?
What is the tension in the wire? What is the net force the hinge exerts on the beam? The maximum tension the wire can have without breaking is
T=936 N. What is the maximum mass sign that can be hung from the beam?
1 answer:
Answer:
wire tension =1003.8N
force in the hinge=894.15N
mass of sign=15.8Kg
Explanation:
Hello!
To solve this problem you must follow the following steps the complete procedure is in the attached images
1. draw the complete free body diagram of the situation
2. Raise Newton's equilibrium equations that establish that for a body to be in equilibrium, the sums of its vertical, horizontal forces and moments with respect to point A must be equal to zero.
3. Use the moment equilibrium equation with respect to point A (see attached image) to find the cable tension.
4. Use the equilibrium equations of vertical and horizontal forces to find the total reaction force on the hinge.
5. Use the equilibrium equation of moments with respect to point A (see attached image) to find the mass of the sign.
Remember to use algebra correctly to find the variables.
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Answer:
A. Part a is the attachment
B. total work = 10.4kj
Explanation:
T1 = constant temperature
nRT1 = PaVa = PbVb
We write equation as
5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)
W1 = 25 ln2
W1 = 25 x 0.693
= 17.327kj
The isochoric expansion has no change in volume. So,
W2 = 0
Isothermal compression
T3 = constant temperature
nRT3 = PcVc = PdVd
Pc = 1mpa Vc = 10L Vd = 5L
= 10x-0.693
= -6.93kj
Isochoric compression has no change in volume. Workdone w4 = 0
Total workdone = w1 + w2 + w3 + w4
= 17.33 + 0 + (-6.93) + 0
= 10.4kj
Answer:
Ф = 2.179 eV
Explanation:
This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.
E = K + Ф
the energy of the photons is given by the Planck relation
E = h f
we substitute
h f = K + Ф
Ф= hf - K
the speed of light is related to wavelength and frequency
c = λ f
f = c /λ
Φ =
let's reduce the energy to the SI system
K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J
calculate
Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹
Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹
Ф = 3.4571 10⁻¹⁹ J
we reduce to eV
Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
Ф = 2.179 eV