A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin
1 answer:
Answer:
To the right relative to the original frame.
Explanation:
In first reference frame <em>S</em>,
Spatial interval of the event,
Temporal interval of the event,
In the second reference frame <em>S'</em>, the two flashes are simultaneous, which means that the temporal interval of the event in this frame is
The speed of the frame <em>S' </em>with respect to frame <em>S</em> = v.
According to the Lorentz transformation,
And positive v means the velocity of the second frame<em> </em><em>S'</em> is along the positive x-axis direction, i.e., to the right direction relative to the original frame <em>S</em>.
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Answer:
1) d = h- L - mg / k , 2) k = 2 mg/h (-1 +2 L / h)
Explanation:
1) Let's use the translational equilibrium equation
-W = 0
F_{e} = W
k x = mg
x = mg / k
from the statement of the exercise the height of the bridge, for the reference system in the river, is
h = L + x + d
x = h - L - d
we substitute
h - L -d = mg / k
d = h- L - mg / k
2) They ask us for the spring constant. For this part we can use energy conservation
Starting point. At the point before jumping
Em₀ = U = m g h
Final Point. When it's hanging
Em_f = + U = ½ k x² + mg d
Em₀ = Em_f
mg h = 1 / k x² + m g d
in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0
mg h = ½ k x²
k = 2 mg h / x²
we substitute the value of x
k = 2mg h / (h -L)²
k = 2mg h / (- L + h)²
we simplify the expression
k = 2mg h / [h² (1- L / h)²]
k = 2m /h (1- L / h)⁻²
In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression
(1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2! (1 -L / h)² + ...
for simplicity let's keep up to the linear term, we substitute in the solution
k = 2 mg/h [1 - 2 (1- L / h)]
k = 2 mg/h (-1 +2 L / h)
Answer:
6 rad/s
Explanation:
In a spring the angular frequency is calculated as follows:
where is the angular frequency,
is the mass of the object in this case
, and
is the constant of the spring.
To calculate the angular frequency, first we need to find the constant which is calculated as follows:
Where is the force:
, and
is the distance from the equilibrium position:
.
Thus the spring constant:
And now we do have everything necessary to calculate the angular frequency:
the angular frequency of the oscillation is 6 rad/s
Answer:
the formula for the force of gravity depends inversely of the distance and directly of the masses.
This means that if the masses increase and their distance remains the same the force will increase proportionally.