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Crazy boy [7]
3 years ago
10

arlene is to walk across a high wire strung horizontally between two buildings 10.0m apart. The sag in the rope when she is at t

he midpoint is 10.0 degrees, if her mass is 50.0 kg, what is the tension in the rope at this point?...can you help me...thanks!
Physics
2 answers:
Ipatiy [6.2K]3 years ago
4 0
We are given with a force in the middle of a rope measuring 10 m and an angle of sagging of 10 degrees from the horizontal. The tension of the rope is equal to the hypotenuse of the right triangle. 

sin 10 = 50 kg * 9.8 m/s2 / T
Tension = 2821. 80 N 
Greeley [361]3 years ago
3 0

Answer:

T = 1412.3 N

Explanation:

As we know that weight of the Arlene is counterbalanced by the tension in the string

So here it is given that string makes 10 degree angle at mid point

So we will have

T sin\theta + T sin\theta = mg

so we have

T = \frac{mg}{2sin\theta}

here we know that

m = 50 kg

\theta = 10 degree

now we have

T = \frac{50 \times 9.81}{2sin10}

T = 1412.3 N

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A liquid is a matter that neither has a fixed shape but fixed volume...in college really need to know this.

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An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?
anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

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t = 10

It takes 10 seconds to land from a height of 250 ft.

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3 years ago
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A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

8 0
3 years ago
At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f
worty [1.4K]

Answer:

v_{min} \approx 17.153\,\frac{m}{s}

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h

The minimum velocity is:

v_{min} = \sqrt{2\cdot g \cdot h}

Let assume that radio of curvature is measured in meters. Hence:

v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}

v_{min} \approx 17.153\,\frac{m}{s}

8 0
3 years ago
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