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Sav [38]
3 years ago
13

A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of

44.0 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 1.37 m/s2?
Physics
1 answer:
vitfil [10]3 years ago
3 0

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

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If a car changes its velocity from 32km/hr in 8.0 seconds what is its acceleration
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We don't know the change in velocity, so can't answer.

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Show the weight of the ladder and draw the missing Frictional force.
madreJ [45]
Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)

So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2

W =  \frac{N}{g}  =  \frac{100N}{9.8m/s^{2}} =  \frac{100}{9.8} = 10.2kg


Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
5 0
2 years ago
A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat
Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

7 0
2 years ago
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