Ask question

Ask question

All categories

3 years ago

10

A 35.0 kg uniform beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizo

ntal. If the angle between the beam and the cable is θ = 59.0° what is the tension in the cable?

1 answer:

Vitek1552 [10]3 years ago

3 0

Answer:

200.077 N

Explanation:

Mass of beam m= 35.0kg

This is a torque problem

Sum of torques should be zero for equilibrium

Marque along point A is

weight force × half legth of beam - T sin (59°) × Length of the beam=0

35.0 × 9.8 × L/2 = T × 0.8571673007 × L

==> 343 /2 = T× 0.8571673007

==> T= 200.077 N

You might be interested in

A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with t

sergij07 [2.7K]

Answer: $135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

<u>*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface</u>

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium refractive index (the air)

$n_{2}=1.309$ is the second medium refractive index (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the <u>angle between the reflected and the refracted light</u>.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$ (2)

Remembering that the Normal makes a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$ (3)

Finding $\beta$ :

$\beta=90\º-25.5\º=64.5\º$ (4)

Doing the same with $\theta_{2}$ and $\alpha}$ :

$90\º=\theta_{2}+\alpha$ (5)

Finding $\alpha$ :

$\alpha=90\º-19.201\º=70.79\º$ (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$ (7)

$\alpha+\beta=135.29\º$ >>>This is the angle between the reflected and the refracted light.

5 0

3 years ago

What do the spheres in the model represent

Sophie [7]

Answer: no

Explanation: we need a picture

4 0

3 years ago

Read 2 more answers

A person walks 2 miles every day to work, leaving her front porch at 7:00 A.M. and arriving at work at 7:30 A.M. ON the way home

Nataly [62]

Answer:

The displacement is zero miles

Explanation:

The displacement of an object that moves from point A to point B is defined as

$d =B-A$

Where d is the displacement of the object. The displacement does not depend on the trajectory of the object. It only depends on the linear distance between the end point and the starting point.

In this case we know that the person walks from home to work and then walks from work to home. Therefore, the total displacement is the linear distance between the point where its journey begins and the point where the route ends.

The tour begins on the front porch of your house and ends on the front porch of your house (when you return from work). If we call A to the front porch of the house then the displacement is:

$d = A-A = 0$

The displacement is zero miles, since the person finishes the journey just where it started (front porch)

7 0

3 years ago

What is the second step in the scientific method

beks73 [17]

The steps<span> of the </span>scientific method<span> are to: Ask a Question. Do Background Research. Construct a Hypothesis. Test Your Hypothesis by Doing an Experiment. Analyze Your Data and Draw a Conclusion.</span>

7 0

4 years ago

Read 2 more answers

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago