The direction of the motion is constantly changing during
motion over any closed path, not only circular.
Answer:This quest took a huge leap forward in 2000 when Hubble studied the exoplanet HD 209458 b, the first extrasolar planet known to make “transits” across the face of its star. Hubble became the first telescope to directly detect an exoplanet's atmosphere and survey its makeup.
Explanation:
The weight of the load is (mass) x (gravity).
The work done to lift it is (weight) x (distance) = (mass) x (gravity) x (distance)
The power is (work)/(time) =
(mass) x (gravity) x (distance) / (time) =
(500 kg) x (9.8 m/s²) x (12 m) / (1 sec) =
(500 x 9.8 x 12 / 1) (kg-m / sec²) (m) / (sec) =
(500 x 9.8 x 12 / 1) newton - meter / sec =
(500 x 9.8 x 12 / 1) joule / sec =
58,800 watts
= <em>58.8 kW</em>
= about 78.8 horsepower
A. The average speed is
<span>12 km/18 min = 0.67 km/min
b. The average velocity is
10.3 km/18 min = 0.57 km/min
c. average speed is 24 km/450 min = 0.05 km/min
average velocity = 0 since the displacement is 0.
</span><span>
</span>
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
<u>F' = F/12</u>
<u>Therefore, the electrostatic force is reduced to one-twelve of its original value.</u>
