Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
Answer:
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun =
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
The gravitational acceleration at the surface of such a star is
The velocity of the object would be 3257806.62409 m/s
<em><u>your </u></em><em><u>question</u></em><em><u>:</u></em><em><u> </u></em>
<em>What are the two types of physical fitness?</em>
<em><u>answer:</u></em>
<em>The</em><em> </em><em>two </em><em>types </em><em>of </em><em>phys</em><em>ical </em><em>fitn</em><em>e</em><em>s</em><em>s</em><em> </em><em>are </em><em>Health-related</em><em> physical fitness and Performance-related physical fitness.</em>