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Flauer [41]
2 years ago
13

The column has a mass of 600 lb/ft and a force of 50 kips applied at the top. The diagonal brace has a 10 kip axial force that c

an be either tension or compression (note that in seismic and wind loading on buildings, braces experience both tension and compression). Assume all elements are pin-connected at their ends, including the two supports.

Physics
1 answer:
drek231 [11]2 years ago
4 0

Answer: the maximum compression force = 59.47 kip

the minimum compression force = 43.33 kip

Explanation:

In this we required to do force balance in order to get maximum and minimum compression force in the column.

the picture below explains the steps to solve the question with diagrams to ease understanding.

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Which of these is a benefit of social networking ?
Alisiya [41]

Answer:

Staying connected to friends

Explanation:

hope this helps

6 0
2 years ago
How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?​
miss Akunina [59]

Answer:

The answer to your question is    Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

           Pe = mgh

Substitution

           Pe = (50)(5)(9.81)

Simplification

           Pe = 2452.5 J

8 0
2 years ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
lisabon 2012 [21]

Answer:

\alpha =10.93radian/sec^2

Explanation:

We have given given the final angular velocity \omega _{final}=13.5rad/sec

And \omega _{initial}=22rad/sec

Displacement \Theta =13.8radian

We have to find the angular acceleration \alpha

According to law of motion \omega _{final}^2=\omega _{initial}^2+2\alpha \Theta

So 13.5^2=22^2+2\times \alpha \times 13.8

\alpha =-10.93radian/sec^2

In question we have tell about magnitude only so \alpha =10.93radian/sec^2

4 0
3 years ago
Need help please... thank u​
tangare [24]

Answer:

the answer from my side is both vertical and horizontal

4 0
2 years ago
Read 2 more answers
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