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t=(0-(250sin75)^2)/-9.8
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>
<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
Answer:
To measure work, you must multiply the force by the distance through which it acts.
Metal conducts heat quicker than wood or air so when you touch it heat is taking away from you body faster than when you touch the wood desk. When heat is taken from our bodies our brain interprets this as cold.
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
