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slega [8]
2 years ago
12

How do arch bridges distribute the load?

Physics
2 answers:
Sveta_85 [38]2 years ago
6 0

Answer:

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

Explanation:

alukav5142 [94]2 years 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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PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST The actual subject is Science but they dont have that as a option in pick
Sophie [7]

Explanation:

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• miles per hour.

• distance/time.

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• miles per hour North

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5 0
2 years ago
Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational
xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

gravitational constant(G)=6.67*10^-11Nm^2kg^-2

gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

•°• F=7.41*10^2

thus, the gravitational force between them is 7.14*10^2

3 0
3 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
What is the displacement of the object from 2 seconds to 6 seconds?
masya89 [10]

the answer would be 4

8 0
3 years ago
Read 2 more answers
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