The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
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s=600 m
t=12 s
s=0.5*a*t² (initial speed V0=0)
a=(2*s)/t²
a=(2*600)/12²
a≈8.33 m/s²
L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)
L=0.5*a*(t2²-t1²)
L=0.5*((2*s)/t²)*(t2²-t1²)
L=0.5*((2*600)/12²)*(12²-11²)
L ≈ 95.83 m
Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is
where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE
We know that capacitrnce and charge is related as
v = 9.523 V
Electric field is given as
= 
E = 5291.00 V/m
E = 5291.00 N/C
Answer:
Covalent Bond
Explanation:
Covalent Bonds happen when elements in a compound share electrons to exist together in a stable state as in the diagram
