The discovery and characterization of cathode rays was important in the development of the atomic theory because
1 answer:
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Solution :
Mass of the particle = M
Speed of travel = v
Energy of one photon after the decay which moves in the positive x direction = 233 MeV
Energy of second photon after the decay which moves in the negative x direction = 21 MeV
Therefore, the total energy after the decay is = 233 + 21
= 254 MeV
So by the law of conservation of energy, we have :
Total energy before the decay = total energy after decay
So, the total relativistic energy of the particle before its decay = 254 MeV
The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
Explanation:
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Answer:
Weight W = 392.4 N
Density = 784.31
Specific gravity S = 0.78431
Force required F = 10 N
Explanation:
Given data
Mass (m) = 40 kg
Volume (V) = 0.051
Weight W = m × g
⇒ W = 40 × 9.81
⇒ W = 392.4 N
This is the weight of the methanol.
Density =
⇒ =
⇒ = 784.31
This is the density of the methanol.
Specific gravity (S) =
⇒
⇒ S = 0.78431
This is the specific gravity of the methanol.
Force needed to accelerate this tank F = ma
⇒ F = 40 × 0.25
⇒ F = 10 N
This is the force required to accelerate the tank.