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deff fn [24]
3 years ago
14

Where are truss bridges located?

Physics
2 answers:
guapka [62]3 years ago
7 0

Basic truss bridge types found in North Carolina (source: HAER) A truss bridge can be characterized by the location of its traffic deck. At a pony truss, the travel surface passes along the bottom chords of trusses standing to either side that are not connected to each other at the top.

pshichka [43]3 years ago
5 0

Where are truss bridges located?

A. North Carolina

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Three-dimensional ____ programs allow designers to rotate designs of 3-d objects to view them from any angle.
notsponge [240]
Long phrase is the blank
5 0
3 years ago
Can someone please help meeeeee please
bagirrra123 [75]

Answer:

0.650

Explanation:

Move the decimal point three times to the left.

3 0
3 years ago
An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0km/hour is blowing southward.
qaws [65]

Answer:

a)  correct answer is C , b) 14º  from the west to the north, c)   v_{1g} = 300.79 km / h

Explanation:

This is a relative speed exercise using the addition of speeds.

1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C

2 and 3) In this case we must compose the speed using the Pythagorean Theorem.

     v_{1a}² = v_{1g}² + v_{ag}²

where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth

in this case let's clear the speed of the airplane with respect to the Earth

  v_{1g} = √(v_{1a}² - v_{ag}²)

 v_{1g} = √ (310² - 75²)

 v_{1g} = 300.79 km / h

we find the direction of the airplane using trigonometry

   sin θ = v_{ag} / v_{1a}

   θ = sin⁻¹ (v_{ag} /v_{1a})

   θ = sin⁻¹ (75/310)

   θ= 14º

the pilot must direct the aircraft at an angle of 14º from the west to the north

7 0
3 years ago
An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period
KatRina [158]

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

   The period of the  asteroid is   T =  176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s

Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

7 0
2 years ago
A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0
3 years ago
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