Answer:
The time taken by missile's clock is
Solution:
As per the question:
Speed of the missile,
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:
(1)
Using binomial theorem in the above eqn:
We know that:
Thus eqn (1) becomes:
Now, putting appropriate values in the above eqn:
Answer:
9.82 × Hz
Explanation:
De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:
λ =
where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.
Given that: h = 6.63 × Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;
λ =
=
=
= 9.8222 ×
The wavelength of the object is 9.82 × Hz.
There are 3 forces acting on the stoplight:
• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward
• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right
The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that
∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0
We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to
2<em>T</em> sin(<em>θ</em>) = <em>W</em>
2 (1000 N) sin(<em>θ</em>) = 100 N
sin(<em>θ</em>) = 0.05
<em>θ</em> ≈ 2.87°
If <em>y</em> is the vertical distance between the stoplight and the ground, then
tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)
Solve for <em>y</em> :
tan(2.87°) = (15 m - <em>y</em>) / (100 m)
<em>y</em> = 15 m - (100 m) tan(2.87°)
<em>y</em> ≈ 9.99 m
