Answer:
1840:1
Explanation:
If m is the mass of the electron, and 1840m is the mass of the proton, then:
p₁ = p₂
m₁v₁ = m₂v₂
m v₁ = 1840m v₂
v₁ = 1840 v₂
The kinetic energy of the electron is:
KE₁ = ½ m₁ v₁²
KE₁ = ½ m (1840 v)²
KE₁ = 1692800 mv²
The kinetic energy of the proton is:
KE₂ = ½ m₂ v₂²
KE₂ = ½ (1840m) v₂²
KE₂ = 920 mv²
The ratio of the kinetic energies is:
KE₁ / KE₂
(1692800 mv²) / (920 mv²)
1840:1
Answer:
Explanation:
Saturn orbital period
p(s)= 29.46years
Average distance
a(s) = 9.54AU
Venus orbital period
p(v) = 0.62 years
a(v) ?
Using Kepler's third law
a³ ∝p²
a³ = kp²
a³/p² = k
Where
a is the distance if planet from sun
T is the period of the planet
Then,
a(Venus)³/p(venus)² = a(saturn)³/p(saturn)²
a(v)³/p(v)² = a(s)³/p(s)²
a(v) ³/ 0.62² = 9.54³/29.46²
a(v) ³/ 0.62² = 1.0004
a(v)³ = 1.0004× 0.62²
a(v)³ = 0.3846
a(v) = cube root(0.3846)
a(v) = 0.727 AU
Answer:
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Explanation:
the fundamental answer is without regular supervision of definition of weights and measures,commerce exchange will be impossible and there would be no market whatsoever for anything.
Water
Air
Heat
Magnetism
Solar power
Answer:
Explanation:
Firs of all, we calculate the position of the image of the ring with the converging lens. After this we take the image of the ring as the object for the diverging lens (we have to use the lens' equation, and we assume that objects to the left are positive and images to the right are positive).
(a) Hence, we have for the first step
so, the image is at a distance of 23.6 cm to the right of the converging lens. The distance between both converging and diverging lens is 19.5cm. Hence the position of the image of the ring (produced by the first lens) is
23.6cm - 19.5cm = 4.1cm
4.1 cm to the right of the diverging lens. Hence we have for a diverging lens
where we take do=-4.1cm beacuse the object is to the right. The image is to the left of the diverging lens
(b)
The magnification is
this constant is a factor for the second magnification
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