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3 years ago

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

Physics

2 answers:

NemiM [27]3 years ago

6 0

Answer:

the sun

Explanation:

because thats it

snow_tiger [21]3 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

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A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago

He conducted experiments in combining elements

Wewaii [24]

Where is the rest .........

3 0

3 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

5 0

3 years ago

If two waves meet and build on each other, blank interference is occurs.

Thepotemich [5.8K]

That is called constructive interference. If they meet and cancel, it is destructive interference

4 0

3 years ago

Read 2 more answers

A 500 kg empty elevator moves with a downwards acceleration of 2 m/s2. What is the tension force of the cable on the elevator? 5

Gre4nikov [31]

Answer:

T = 4000 N

Explanation:

We have,

Mass of an elevator, m = 500 kg

It is moving downward with an acceleration of $2\ m/s^2$ . It is required to find the tension force of the cable on the elevator. We know that when the elevator is moving downward, the net force on it is given by :

$F=m(g-a)$

g is acceleration due to gravity

$F=500\times (10-2)\\\\F=4000\ N$

So, the tension force of the cable on the elevator is 4000 N.

5 0

3 years ago