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
Explanation:
(a)
Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.
Using Snell's law as:
Where,
is the angle of incidence
is the angle of refraction = 90°
is the refractive index of the refraction medium
is the refractive index of the incidence medium
Thus,
The formula for the calculation of critical angle is:
Where,
is the critical angle
(b)
No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.
Answer:
The answer to the question is as follows
The acceleration due to gravity for low for orbit is 9.231 m/s²
Explanation:
The gravitational force is given as
Where 
= Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by and = 1 kg
rEarth = The distance between the two masses = 6.40 x 10⁶ m
therefore at a height of 400 km above the erth we have
r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m
and = 
= 9.231 N
Therefore the acceleration due to gravity = /mass
9.231/1 or 9.231 m/s²
Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²
The approximate lateral area of the prism is determined as 831 square inches.
<h3>
What is lateral area of the hexagonal prism?</h3>
The lateral area of the hexagonal prism is calculated as follows;
LA = PH
where;
- P is perimeter of the prism
- H is height
A = ¹/₂Pa
where;
- a is apothem = 10 inches
- A is base area = 346.41 in²
346.41 = ¹/₂(10)P
346.41 = 5P
P = 346.41/5
P = 69.282 inches
LA = PH
LA = 69.282 x 12
LA = 831.38 in²
Learn more about lateral area of prism here: brainly.com/question/296674
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