All categories

2 years ago

What is the primary force help suspension bridges use cables to hold there spans up

Physics

2 answers:

NemiM [27]2 years ago

6 0

Answer:

the sun

Explanation:

because thats it

snow_tiger [21]2 years ago

3 0

Explanation:

Suspension bridges, like the Golden Gate Bridge or the Brooklyn Bridge, use tension force as the primary source of force that cables use to hold their spans up. The supporting cables receive the tension forces of the bridge, and this same force passes to the anchorages and into the ground

You might be interested in

A students walks one hour. She walks 4 blocks east and 2 blocks west. Her average velocity is?

Rama09 [41]

Let us consider the following figure

figure is in picture

So, his/her total displacement = 2 blocks.

total time = 1 hour

Average Velocity

= Total displacement/ total time

= 2 blocks/ 1 hour

= 2 blocks/ h

So, her/his average velocity is 2 blocks per hour.

6 0

3 years ago

The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}

tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

$K.E=\frac{1}{2}mv^2$

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2$

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

$K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2$

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

$\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}$

Given that, v = 2v'

$\frac{K.E}{K.E'}=\frac{1}{1}$

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0

3 years ago

Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

3 years ago

A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance

FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0

3 years ago

I need 4 answers for Earth's layers and 4 answers for the Descriptions for Earth's layers graphic organizer.

Arada [10]

Ok well have a nice day

3 0

3 years ago