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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:

You might be interested in
Formula:
s2008m [1.1K]

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

8 0
3 years ago
You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through
Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0
3 years ago
How much work is done when a 100 N force moves a block 59 m?
xxTIMURxx [149]

Answer:

5900J

Explanation:

Work=Forse*Distance

work = J, Jewls

100*59=5900

Hop this helps and can u think about brainlist

i put a picture on how to find these answers, if u got any more questions im here

3 0
3 years ago
Which type of exercise is weightlifting?
MatroZZZ [7]

The answer is option B "anaerobic." Weightlifting deals with stress to the muscles when lifting weights and due time the muscles will begin to adapt and get stronger. Other examples of anaerobic exercise are things like: weight training, sprinting, cycling, and jumping anything that has short exertion, and high-intensity movement is an anaerobic exercise.

Hope this helps!

Nonportrit

5 0
3 years ago
Read 2 more answers
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
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