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gtnhenbr [62]
3 years ago
15

A fisherman is fishing from a bridge and is using a "40.0-N test line." In other words, the line will sustain a maximum force of

40.0 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 2.57 m/s2?
Physics
1 answer:
IRISSAK [1]3 years ago
7 0

To solve this problem we require the concept of Forces given by Newton in his second law. For the particular case where the acceleration is 0 and when there is a change of acceleration.

A) For the first case where the speed is constant we have to

F = mg

Here,

a=0m/s^2

Then,

F= mg

40N = m (9.8)

m = 4.077kg

B) Here we have an acceleration of 2.57m / s ^ 2. When there is acceleration it is considered that the sum of forces is dynamic, for that reason when performing the summation, we have

\sum F = ma

F-mg = ma

T = m(a+g)

40= m(9.8+2.57)

m = 3.23kG

Therefore the heaviest fish that can be pulled up vertically is 3.23KG when there is acceleration but 4.077kg with constant velocity

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Answer:

Explanation:

Two straight wires

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i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

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3 years ago
The CIA investigated hypnosis as a possible tool for interrogating prisoners. Why did they decide it was unsuitable for that
alukav5142 [94]

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D. Hypnosis can make the subjects talk, but they talk only about their childhoods.

Explanation:

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3 years ago
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How am I supposed to solve this?
RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

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The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
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Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

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Answer:

1

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