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jek_recluse [69]
3 years ago
11

What is the primary force that helps suspension bridges use cables to hold their spans up? A. tension force B. resistance force

C. normal force D. elastic force E. applied force
Physics
2 answers:
dlinn [17]3 years ago
7 0

Answer:

The correct answer is option A. "tension force".

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. This equilibrium that the tension force gives among the different elements of the bridge, gives the support to hold the bridge together.

aleksandr82 [10.1K]3 years ago
5 0

Answer:

A. Tension force

Explanation:

These are bridges that use cables to hold the road way up.

The cables in a suspension bridge receive the tension force of the bridge. These cables transmit these forces to the towers supporting the bridge and are compressed into the ground.

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How to find the frictional force acting on an object (not the friction coefficient)? ...?
Dovator [93]
One can simply find the frictional force acting on an object using this equation:

 (Ffrict<span> = μ•F</span>norm<span>)
</span>

The process of determining the value of the individual forces acting upon an object involve an application of Newton's second law (Fnet=m•a) and an application of the meaning of the net force. If mass (m) and acceleration (a) are known, then the net force (Fnet) can be determined by use of the equation.

<span>Fnet = m • a</span>

If the numerical value for the net force and the direction of the net force is known, then the value of all individual forces can be determined.

7 0
3 years ago
Answer the question plz
ollegr [7]

Answer:

b and d

Explanation:

7 0
3 years ago
At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?
Mrac [35]

Answer:

The value of x is 2.1 cm from the center of the coil.

Explanation:

Radius, R = 2.7 cm

Number of turns, N = 800

The magnetic field at the axis is half of the magnetic field at the center.

B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2  \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm

4 0
3 years ago
SI Prefix Meaning
LenKa [72]
- deci: one-tenth.
- centi: one-hundredth.
- nano: one-billionth.

hope this helps :)
7 0
3 years ago
An archer fires and arrow while standing atop a 5.15 m tall wall. The arrow is fired at an angle of 55 degrees and has a launch
il63 [147K]

Answer:

Explanation:

Given

height of wall=5.15 m

angle of launch(\theta )=55^{\circ}

Launch velocity(u)=52.4 m/s

Time of flight will be sum of time of flight of projectile+time to cover 5.15 m

Time of flight of arrow=\frac{2usin\theta }{g}

t=\frac{2\times 52.4\times sin55}{9.81}=8.76 s

Now time require to cover 5.15 m

Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.

v_y=-52.4sin55 at zero vertical displacement

Thus time required will be t_2

5.15=52.4sin55\times t+\frac{gt^2}{2}

4.9t^2+42.92t-5.15=0

t=0.118 s i.e. t_2=0.118 s

total time =t_1+t_2=8.76+0.118=8.878 s

(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s(R_2)

R_1=\frac{u^2sin2\theta }{g}=\frac{52.4^2\times sin110}{9.8}=263.28 m

R_2=52.4cos55 \times 0.118=3.546 m

R=R_1+R_2=263.28+3.546=266.82 m

8 0
3 years ago
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