We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2). <span>
We want a ratio of 3000000 to 1, or 3 million to 1.
</span>
<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
Answer:
Explanation:
We shall apply Pascal's Law in fluid mechanics
According to it , pressure is transmitted in liquid from one point to another without any change .
25 cm diameter = 12.5 x 10⁻² m radius
Area = 3.14 x (12.5 x 10⁻²)²
= 490.625 x 10⁻⁴ m²
Pressure by vehicle
Force / area
13000 / 490.625 x 10⁻⁴
= 26.497 x 10⁴ Pa
5 cm diameter = 2.5 x 10⁻² radius
area = 3.14 x (2.5 x 10⁻²)²
= 19.625 x 10⁻⁴ m²
If we assume required force F on this area
Pressure = F / 19.625 x 10⁻⁴ Pa
According to Pascal Law
F / 19.625 x 10⁻⁴ = 26.497 x 10⁴
F = 19.625 x 26.497
= 520 N
Answer:
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Explanation:
