Answer:
The distance traveled by the woman is 34.1m
Explanation:
Given
The initial height of the cliff
yo = 45m final, positition y = 0m bottom of the cliff
y = yo + ut -1/2gt²
u = 20.0m/s initial speed
g = 9.80m/s²
0 = 45.0 + 20×t –1/2×9.8×t²
0 = 45 +20t –4.9t²
Solving quadratically or by using a calculator,
t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s
So this is the total time it takes for the ball to reach the ground from the height it was thrown.
The distance traveled by the woman is
s = vt
Given the speed of the woman v = 6.00m/s
Therefore
s = 6.00×5.69 = 34.14m
Approximately 34.1m to 3 significant figures.
Answer:
Total distance travelled = 210m
Explanation:
Distance travelled = 80m + 50m + 10m + 70m
= 210m
Answer:
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You have to put the work to it and but the answer is 30cm
d = distance between the two point charges = 60 cm = 0.60 m
r = distance of the location of point "a" where the electric field is zero from charge 
between the two charges.
= magnitude of charge on one charge
= magnitude of charge on other charge
= 3
= Electric field by charge at point "a"
= Electric field by charge at point "a"
Electric field by charge at point "a" is given as
= k /r²
Electric field by charge at point "a" is given as
= k /(d-r)²
For the electric field to be zero at point "a"
=
k /(d-r)² = k /r²
/(d-r)² = 3 /r²
1/(0.60 - r)² = 3 /r²
r = 0.38 m
r = 38 cm
