When is a train on Earth in motion relative to the sun? A It is relative to the sun only when the train is not moving. B It is n
1 answer:
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Answer:
dJ = 1.7 m
Explanation:
The Equation of the Balancing the moments in the center of the seesaw is like this:
∑Mo = 0
Mo = F*d
Where:
∑Mo : Algebraic sum of moments in the center(o) of the balance
Mo : moment in the o point ( N*m)
F : Force ( N)
d : distancia of the force to the the o point ( N*m)
Data
mA = 60 kg : mass of the Anna
mJ = 70 kg : mass of theJon
dA = 2 m : Distance from Anna to the center of the seesaw
g: acceleration due to gravity
Calculation of the distance from Jon to the center of the seesaw (dJ)
∑Mo = 0 WA : Ana's weight , WJ : Jon's weight
W = m*g
(WA)(dA) - (WJ) (dJ) = 0
(mA*g)(dA) - (mJ*g)(dJ) = 0
We divide by g the equation:
(mA)(dA) - (mJ)(dJ)= 0
(mA)(dA) = (mJ)(dJ)
dJ = 1.7 m
Answer:
option A
Explanation:
given,
height of the drop of stone = 9.44 m
speed of the stone = ?
As the stone is dropped the energy of the stone will be conserved.
using conservation of energy.
Potential energy = Kinetic energy
v = 13.60 m/s
Hence, the correct answer is option A
Answer:
the electric field at point A is
E = 5.5 ×10¹³N/C(-x direction)
Explanation:
given
electrostatics constant k = 9.0×10⁹
charge q = 31.7mC= 31.7×10⁻³C
distance r = 2.80mm
distance from midpoint to point A = 2.00mm
attached is the diagram of the solution, describing the position of the charge
note x = r/2, where x is the distance from midpoint of r to the particle
using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m
the electric field at point A is given as
vector <em>E </em>= 2E×cos θ( -x direction)
recall E =kq/x²
where k is the electrostatics constant = 1/4πε₀
where ε₀ is permittivity of free space
therefore using E =2{kq/x²}cosθ
∴cosθ = adjacent/hypotenuse
cosθ=1.40/2.44
E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)
E=2{4.79×10¹³}×(0.574)(-x)
E = 2×2.75 ×10¹³N/C(-x direction)
Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)
