Neither set of choices is correct.
If the distance is tripled, then the forces decrease to
1/9 Fg. and. 1/9 Fe.
Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values
ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.
Answer:
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Explanation:
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Answer:
0.0109 m ≈ 10.9 mm
Explanation:
proton speed = 1 * 10^6 m/s
radius in which the proton moves = 20 m
<u>determine the radius of the circle in which an electron would move </u>
we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)
For proton :
Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )
For electron:
re = Me*V/ qE * B -------- ( 2 )
Next: take the ratio of equations 1 and 2
re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )
∴ re ( radius of the electron orbit )
= ( Me / Mp ) rp
= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20
= ( 5.45 * 10^-4 ) * 20
= 0.0109 m ≈ 10.9 mm
