How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would
1 answer:
Answer: 1037 miles per hour
Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.
Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,
Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by
Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.
The obtained result is 1037 miles per hour.
You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.
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Answer:
a) T_A = , b) T_B = g [m₂ (
) + m₁ (
) ]
c) x = , d) m₂ = m₁ (
)
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A =
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = - g m₂ - g m₁
T_B = g [m₂ ( ) + m₁ (
) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x =
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
m₂ = m₁
m₂ = m₁ ( )
