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slega [8]
1 year ago
12

How do arch bridges distribute the load?

Physics
2 answers:
Sveta_85 [38]1 year ago
6 0

Answer:

carried outward along the curve of the arch to the supports at each end

Explanation:

alukav5142 [94]1 year ago
3 0

Answer:

Instead of pushing straight down, the load of an arch bridge is carried outward along the curve of the arch to the supports at each end. The weight is transferred to the supports at either end. These supports, called the abutments, carry the load and keep the ends of the bridge from spreading out.

Explanation:

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A man attempts to push a 19.8 kg crate across a warehouse floor. He slowly increases the force until the crate starts to move at
VARVARA [1.3K]

Answer:

 μ = 0.18

Explanation:

Let's use Newton's second Law, the coordinate system is horizontal and vertical

Before starting to move the box

Y axis

     N-W = 0

     N = W = mg

X axis

     F -fr = 0

     F = fr

The friction force has the formula

     fr = μ N

     fr =  μ m g

At the limit point just before starting the movement

     F = μ m g

     μ = F / m g

calculate

      μ = 34.8 / (19.8 9.8)

    μ = 0.18

7 0
2 years ago
Read 2 more answers
The value of acceleration due to gravity is less at the top of Mount Everest then that in the terai reason, why?​
xxTIMURxx [149]

Answer:

The value of acceleration due to gravity is greater in terai than in mountain. In terai region the radius of earth is less as it lies close to the centre of the earth. Thus, the value of g is more in terai region.

3 0
2 years ago
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
3 years ago
The simulation shows a game of tug-of-war. Place a blue team and a red team with same-sized people on each side of the rope. Wha
Leona [35]

Answer:

When same-sized team members are placed on each side of the rope, the sizes of the arrows on both sides remain the same.

Explanation:

This is the answer on Plato

6 0
3 years ago
Read 2 more answers
Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0
3 years ago
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