Answer:
A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.
Explanation:
τchild=τrock
We will utilize the formula for torque:
(F)child(d)child)=(F)rock(d)rock)
The gravitational force acts equally on both objects.
(m)childg(d)child)=(m)rockg(d)rock)
We can eliminate gravity from both sides of the equation for simplification.
(m)child(d)child)=(m)rock(d)rock)
Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.
(25kg)(1m)=(50kg)drock
Solve for the distance where the rock should be positioned in relation to the seesaw's center.
drock=25kg⋅m50kg
drock=0.5m
Answer:
I do not communicate in Spanish
Explanation:
To tackle this question, we know the following:
1 Albert equals 88 meters.
1 A = 88 m.
Initially, we square both sides of the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Thus, one acre is approximately 0.52 square alberts.</span>
Answer:3.87*10^-4
Explanation:
To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.
From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.
Solving gives frequency F=0.87*10^20 Hz.
Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.
Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).
This energy is then converted from joules to MeV.
Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding
3.87*10^-4 u.
J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
