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Gnom [1K]
3 years ago
6

A large fish hangs from a spring balance supported from the roof of an elevator. If the elevator has an upward acceleration of 2

.45 m/s
2 and the balance reads 60.0 N, what is the true weight of the fish?
Physics
1 answer:
inn [45]3 years ago
6 0

Answer:

Original Weight = 48 N

Explanation:

When an object hanging from the roof of an elevator and the elevator moves upward with a certain acceleration say a, then the observe weight is always greater than the actual weight.

And when elevator moves downward direction with acceleration a, then the observe weight is always lesser than the actual weight.

Given data:

a = 2.45 m/s²

Balance reading = Tension = T = 60 N

As elevator moves upward,

So,

               T = m(g + a)

         60 N = m(9.81 + 2.45)

               m = 60/12.26

               m = 4.89 kg

This is the original mass of the hanging fish.

We know that

                Weight = W = mg

                                W = 4.89 × 9.81

                                W = 48 N

       

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m
BARSIC [14]

A) The resultant force is 30.4 N at 25.3^{\circ}

B) The resultant force is 18.7 N at 43.9^{\circ}

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

F_1 = 20 N at 0^{\circ} above x-axis

F_2 = 15 N at 60^{\circ} above y-axis

Resolving each force:

F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N

F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N

So, the components of the resultant are:

F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

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F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N

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\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

So, the components of the resultant this time are:

F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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