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Igoryamba
2 years ago
8

Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t

he lowest point of her swing. tension is a vector directed along the chain, measured in newtons, abbreviated n. what are the horizontal and vertical components of the tension at this point in the swing

Physics
1 answer:
Leno4ka [110]2 years ago
8 0
Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°

From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.

The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.

Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.

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the thickness required of a masonry wall L = 375mm

Explanation:

The detailed steps and appropriate use of fourier's law of heat conduction is as shown in the attached file.

5 0
2 years ago
A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p
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Answer:

Explanation:

Given that,

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4 0
2 years ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
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To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

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From our given data we know that,

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\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

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Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

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\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

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4 0
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a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

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Now let's use Newton's second law in the mass that descends

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We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

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b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

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6 0
2 years ago
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