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3 years ago

A 30 kg buzzard is sitting on a cable. The cable is attached to two poles 25

Physics

1 answer:

sertanlavr [38]3 years ago

6 0

Answer:

T1 = 675 [N] ; T2 = 707.5 [N]

Explanation:

To solve this problem we must draw a free body diagram, where the bird is located, the posts and distances mentioned.

In the attached image we can see the free body diagram with different forces, angles and distances.

By conducting a distance analysis we can find the angles on both sides of the rope relative to the horizontal.

Then we perform a force analysis on the X-axis and the y-axis in order to find the two equations we need to find the stresses in the cables (T1 & T2)

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The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of

Fed [463]

Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is:

$\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)$

$\Delta l = 0.02\,m$

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

4 0

3 years ago

A heliocentric system is _____-centered.

aniked [119]

A heliocentric system is a sun-centered

3 0

4 years ago

Read 2 more answers

A soccer ball of mass 0.35 kg is rolling with velocity 0, 0, 1.8 m/s, when you kick it. Your kick delivers an impulse of magnitu

Mademuasel [1]

Answer:

position 9.58 m

Explanation:

In impulse exercises and amount of movement, we always assume that the contact time is small,

I = Δp

With this expression we can calculate the final speed

I = m Vf - m Vo

Vf = (I + mVo) / m

Vf = (1.8 + 0.35 1.8) /0.35

Vf = 6.94 m / s

To calculate the acceleration of the ball we use Newton's second law, after finishing the impulse

∑ F = m a

fr = m a

a = fr / m

a = -0.26 / 0.35

a = -0.74 m/s²

A negative sign indicates that this acceleration is slowing the ball

Now we have speed and time acceleration, so we can use the kinematic equations to find the position at 1.5 s

X = Vo t + ½ to t²

In this case Vo is the speed with which the ball comes out after the impulse 6.94

X = 6.94 1.5 + ½ (-0.74) 1.522

X = 9.58 m

7 0

3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th

gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: $m_1x_1=m_2x_2$

$25 kg\times x=37 kg\times (1.90-x)$

$x=1.1338 m$

The center of gravity is 1.1338 m away from the left side of the barbell

7 0

4 years ago

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What is the frequency of motion of a 0.50 m long pendulum?

IgorC [24]

Complicated 2-step process.

1. Write down the formula for "frequency of a pendulum" from your textbook. 

f = (1/2pi) * sqrt(g/L) 

2. Plug in g and L.

I hope this can help!

3 0

3 years ago

Read 2 more answers