1. Un cable está tendido sobre dos postes colocados con una separación de 10,0 m. A la
1 answer:
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Answer:
0.119 s
Explanation:
Given that
Lets take final speed of the softball after covering 4.8 m = V
We know that
Also We know that
Putting the value of V ,U and a in the previous equation We get
Therefore slide time will be 0.119 s
Answer:
7.65x10^3 m/s
Explanation:
The computation of the satellite's orbital speed is shown below:
Given that
Earth mass, M_e = 5.97 × 10^24 kg
Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg
Orbital radius, r = 6.80 × 10^6m
Based on the above information
the satellite's orbital speed is
V_o = √GM_e ÷ √r
= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6
= 7.65x10^3 m/s
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip