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Over [174]
2 years ago
12

A uniform 1.6-kg rod that is 0.89 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs han

g straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 31 N/m and 63 N/m. Find the angle that the rod makes with the horizontal.
Physics
1 answer:
artcher [175]2 years ago
4 0

Answer:

8.27°

Explanation:

To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.

d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m

by a simple trigonometric relation you obtain that the angle:

sin\theta=\frac{d}{l}=\frac{0.128m}{0.89m}=0.144\\\\\theta=sin^{-1}(0.144)=8.27\°

hence, the angle between the rod and the horizontal is 8.27°

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3 years ago
Read 2 more answers
A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien
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Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

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Solve for a=4.19m/s^2

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3 years ago
Convection currents in air and water occur because _
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2 years ago
A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter
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Answer: 56.87m/s^{2}

Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

F_{net}=F_{up}-W  (1)

Where:

F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

W=m.g=(3kg)(9.79m/s^{2}) (2)

W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

F_{net}=m.a  (6)

Finding the acceleration a:

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a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

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