speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
Learn more about equations of motion from here,
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A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.
Answer:
A) F = V²E_o•πr²/2d²
B) U = E_o•Aπr²V²/2d
Explanation:
A) Since we have two circular plates, the formula for the electric field is expressed as;
E = V/d
Where;
V is voltage
d is distance
However, the net electric field produced is given by;
E' = V/2d
The tension in the cable can then be expressed as;
F = qE'
Where q is charge
Thus;
F = qV/2d - - - (eq 1)
We also know that;
C = q/V = E_o•A/d
A is area = πr²
Thus;
q/V = E_o•πr²/d
q = VE_o•πr²/d
Let's put VE_o•πr²/d for q in eq 1 to get;
F = V²E_o•πr²/2d²
B) formula for the energy stored in the electric field is;
U = ½CV²
From earlier, we saw that; C = E_o•A/d
Thus;
U = ½E_o•AV²/d
A = πr²
Thus;
U = E_o•Aπr²V²/2d
Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV