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sergejj [24]
1 year ago
11

The free-body diagram below represents the force of several vehicles driving across a bridge. Assume that the bridge is in stati

c equilibrium and that it has zero weight, and solve for the unknown reaction force.
Physics
1 answer:
krek1111 [17]1 year ago
7 0

In order to calculate the unknown reaction force, we need to know that the sum of forces pointing down is equal the sum of the forces pointing up, so all forces will be in equilibrium.

So we have:

\begin{gathered} 17800+150000+13200=121000+x \\ 181000=121000+x \\ x=181000-121000 \\ x=60000 \end{gathered}

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I NEED HELP WITH THIS PLEASE
Ann [662]

Answer:

D

Explanation:

One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.

6 0
3 years ago
What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?
ratelena [41]
= 80 N/40 kg
= 2 m/s 2
4 0
3 years ago
An automobile rounds a curve of radius 50.0 m on a flat road.
bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve  = 50m

Centripetal acceleration  = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

 It is given as;

      a = \frac{v^{2} }{r}  

a is the centripetal acceleration

v is the speed

r is the radius

  Now insert the parameters and find v;

         v²   = ar

        v² = 3.92 x 50  = 196

         v  = √196 = 14m/s

6 0
3 years ago
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
A 53.3 kg woman slides down a 35.0° hill with an acceleration of 4.10 m/s. What is the friction force acting on the woman?
lorasvet [3.4K]

Answer:

I attached an image that should help.

Explanation:

Check it out.

5 0
2 years ago
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