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uysha [10]
3 years ago
8

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

42.0 N without breaking. What is the weight of the heaviest fish that can be pulled up vertically, when the line is reeled in (a) at constant speed and (b) with an acceleration whose magnitude is 1.41 m/s2?
Physics
1 answer:
lara31 [8.8K]3 years ago
8 0

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

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A 2500-ohm is connected to a 110v power supply. What is the current through the resistor
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a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec
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Answer: 49.5 m

Explanation:

The speed of sound s is given by a relation between the distance d and the time t:

s=\frac{d}{t} (1)

Where:

s=330 m/s is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

d=412.5 m-x since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance x towards the cliff

t=\frac{2.2 s}{2}=1.1 s Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate d and then find x:

d=st (2)

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3 years ago
A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m
belka [17]

Answer:

Energy stored in the capacitor is U=2.7\times 10^{-5}\ J        

Explanation:

It is given that,

Charge, q=1.5\ \mu C=1.5\times 10^{-6}\ C

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U=\dfrac{1}{2}q\times V

U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36  

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So, the potential energy is stored in the capacitor is U=2.7\times 10^{-5}\ J. Hence, this is the required solution.

4 0
3 years ago
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