A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.
1 answer:
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The correct answer to the question is : 29.88 m.
EXPLANATION :
As per the question, the mass of the rock m = 50 Kg.
The rock is rolling off the edges of the cliff.
The final velocity of the rock when it hits the ground v = 24 .2 m/s.
Let the height of the cliff is h.
The potential energy gained by the rock at the top of the cliff = mgh.
Here, g is known as acceleration due to gravity, and g =
When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.
When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.
The kinetic energy of the rock when it touches the ground is given as -
Kinetic energy K.E = .
From above we know that -
Kinetic energy at the bottom of the cliff = potential energy at a height h
⇒
⇒
⇒
⇒
Hence, the height of the cliff is 29.88 m
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer:
1. = 5.45 m/s
, 2. K = 326.73 J and 3. h = 152 cm
Explanation:
R1. Let's use the conservation of the moment, for this we define a system formed by the two bodies, the pill plus the hanging mass,
Where the mass of the tablet (m = 2 kg) and the hanging mass (M = 38 Kg)
Initial, before crash
po = m v₀₁ + 0
Final, just after the crash
= (m + M)
The moment is preserved
p₀ =
m v1o = (m + M)
= m / (m + M) v1o
= 2/(2+20) 60
= 5.45 m/s
R2 The kinetic energy is given, in our case, after the collision
K = ½ (m + M) ²
K = ½ (2 +20) 5.45²
K = 326.73 J
R3 Let's use the conservation of mechanical energy, after the crash. Let's look for energy at two points the lowest and the highest point
Lowest point
Em₀ = K = ½ (m + M) ²
Highest point
= U = mg h
Em₀ =
½ (m + M) ² = (m + M) g h
h =² / 2g
h = 5.45²/2 9.8
h = 1.52 m (100cm / 1m)
h = 152 cm