1. farther
2. true
the equator is a weirdo
Answer:
Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.
Explanation:
Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 × 
= 0.02 x (180/
)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ = 
So that,
Tan 1.146° = 
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.
Answer:
38.87 m/s
Explanation:
Given that the ball is dropped from a height = 77 m
u = 0 m/s
s = 77 m
a = g = 9.81 m/s²
Applying the expression as:
Applying values as:
<u>The speed with which the ball hit the ground = 38.87 m/s</u>
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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