The answer to the question is sound
Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N
Slow-twitch<span> muscles help enable long-endurance feats such as distance running.
</span>fast-twitch<span> muscles fatigue </span>faster<span> but are used in powerful bursts of movements like sprinting.</span>
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s
Answer:
25N
Explanation:
Assuming the lab is on earth:
w = mg = 2.5 (9.81) = 25N
