Answer:
An apple in free fall accelerates toward the Earth with a free fall acceleration, g. The force of the apple on the Earth also causes the Earth to accelerate toward the falling apple. By Newton's Third Law, the force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth. By Newton,s Second law, the force of the Earth on the apple is equal to the mass of the apple times g , the accelerations due to gravity. And, the force of the the apple on the Earth is equal to the mass of the Earth times the acceleration of the Earth toward the apple. In conclusion, the magnitude of the forces are equal, or
F ( apple on the Earth) = F( the Earth on the apple) or
M( mass of the earth) x a( the acceleration of the earth toward the apple) = m(mass of the apple) x g( the acceleration of the apple toward the Earth) or
a = (m/M) g
Explanation:
For a vertical spring launcher is attached to the top of a block and a ball is placed in the launcher, the position of the ball will be behind the box
<h3>What will be the position of the ball relative to the spring launcher?</h3>
Generally, the equation for the conservation of momentum principle is mathematically given as
(M+m) V1 = M*V2
Therefore, with the ball moving forward we have that; the ball at top it wii be behind the box,
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Answer:
Explanation:
Given that
speed u=4×10⁶ m/s
electric field E=4×10² N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4×10⁶[2×2×10⁻²/7×10¹³]^1/2
=9.5cm
now we find the velocity f the electron strike the plate
v²-(4×10⁶)²=2×7×10¹³×2×10⁻²
v²=16×10¹²+28×10¹¹
v²=1.88×10¹³m/s
speed after hits =>V=4.34×10⁶ m/s
A probability contour represents a bounded, finite volume about a nucleus in which there is a substantial probability of finding the electron is TRUE.
The rectangular of the wave characteristic, ψ2 , represents the opportunity of locating an electron in a given vicinity within the atom. probability finding electron at nucleus is zero.
An orbital is a mathematical feature that has a fee at all points in space. The magnitude squared of that fee is the chance according to volume of locating an electron in that volume of area.
The probability of locating electron in nodes is 0, but there are electron densities on all round nodes. The node is a point or a floor (relying at the type of node) so the volume of the vicinity wherein ψ=zero is zero. we want to position V=zero and we get P=zero so the chance of finding the electron on the node is zero.
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