F = ma
F = (6kg) (3.5m/s2)
F = 21 N
Q = mass water x specific heat water x delta T.
<span>714,000 = mass water x specific heat water x 30.
Substitute specific heat water and solve for mass water.</span>
Answer:
41.0 m/s, 10.6 s
Explanation:
Given:
a = -4.8 m/s²
Δy = 165 m
v = -10.0 m/s
Find: v₀ and t
v² = v₀² + 2aΔy
(-10.0 m/s)² = v₀² + 2(-4.8 m/s²) (165 m)
v₀ = 41.0 m/s
Δy = vt − ½ at²
165 m = (-10.0 m/s) t − ½ (-4.8 m/s²) t²
165 = -10t + 2.4t²
0 = 2.4t² − 10t − 165
Solve with quadratic formula:
t = [ -(-10) ± √((-10)² − 4(2.4)(-165)) ] / 2(2.4)
t = (10 ± √1684) / 4.8
t = 10.6 s
Answer:
1.29 s
Explanation:
Given:
Δy = 1.40 m
v₀ = 0 m/s
a = 1.67 m/s²
Find: t
Δy = v₀ t + ½ at²
(1.40 m) = (0 m/s) t + ½ (1.67 m/s²) t²
t = 1.29 s
Answer:A 0.187 A current flows through a wire. How much time will it take for 2.00 C of charge to flow past a point in the wire?
A. 10.7 s
B. 2.18 s
C. 0.374 s
D. 0.0935 s
Explanation:
A 0.187 A current flows through a wire. How much time will it take for 2.00 C of charge to flow past a point in the wire?
A. 10.7 s
B. 2.18 s
C. 0.374 s
D. 0.0935 s
