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Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
⇒ E = 1.04*10⁻¹ N/C
Answer:
a. -6.17 rad
Explanation:
60 seconds is 2π radians. Writing a proportion:
2π / 60 = x / 59
x = 6.17
The displacement is negative because the second hand moves clockwise.
Answer: D <u>(chemical</u> -> <u>heat</u> -> <u>mechanical</u>)
In automobile engines the petrol/diesel fuel enter in to the engine cylinder, due to spark at the end of the compression, fuel burnt increase the temperature and pressure, develops heat <em>(chemical energy -> heat energy). </em><em>This heat energy acts on a piston develops the work on the crankshaft </em><em>( Heat energy -> Mechanical energy)</em><em>. </em>
