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elena-14-01-66 [18.8K]
2 years ago
10

When engineers consider the forces that can act on a structure, they must think about all parts of the structure. Describe three

different places on a bridge that would experience very different forces.
Physics
1 answer:
REY [17]2 years ago
7 0

Any bridge is subject to three different types of forces: the dead load, the live load, and the dynamic load. The weight of the bridge itself is discussed in the first of these words.

<h3>What are the 4 different types of forces?</h3>

Any of the four fundamental forces in physics—gravitational, electromagnetic, strong, and weak—that control how objects or particles interact as well as how some particles decay—is referred to as a fundamental force, also known as a fundamental interaction.

<h3>What primary force is acting on the bridge?</h3>

Compression and tension are the two main forces that are ever-present on a bridge. A force that compresses or shortens the object it is acting on is known as a compressive force. A force known as tension or tensile force works to stretch or expand the object it is acting on.

To know more about Force visit:

brainly.com/question/13191643

#SPJ13

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A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration
Oksi-84 [34.3K]

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

7 0
3 years ago
A car slams on the brakes to stop. The road pushes back up on the car with a normal force of 12,750 N and friction from the brak
ivann1987 [24]

Answer:

\displaystyle \mu_d=0.75

Explanation:

Coefficients of Friction

Objects in physical contact produce friction which usually manifests as thermal energy being dissipated in the surface where the objects are interacting. It's usually harder to start to move an object from rest, that keeps moving it at a constant speed on the same surface. That is why there are two different coefficients of friction: the static and the dynamic. As mentioned, the static coefficient \mu_s is greater than the dynamic coefficient \mu_d. The car is already moving and is attempting to stop. The coefficient of friction is defined as

\displaystyle \mu_d=\frac{F_r}{N}

Where Fr is the force of friction and N is the normal or the force the road pushes back up on the car. With the given data, we have

\displaystyle \mu_d=\frac{9,560\ N}{12,750\ N}

\displaystyle \boxed{\mu_d=0.75}

The coefficient of friction is dimensionless (doesn't have any units)

6 0
3 years ago
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second
vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

I' = \frac{I_0}{2} cos^2 \theta

Where

\theta =Angle From vertical of the axis of the polarizing filter

I_0 = Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

I = \frac{I_0}{2}

Replacing we have that

I = \frac{370}{2}

I = 185W/m^2

Re-arrange the equation,

I'= \frac{I_0}{2}cos^2\theta

Re-arrange to find \theta

cos^2\theta = \frac{2I'}{I_0}

cos^2\theta = \frac{2*138}{370}

\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})

\theta = 0.5282rad

\theta = 30.27\°

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0
3 years ago
How long does a ship take to cross the pacific?
lidiya [134]
Between 2 weeks and a month

4 0
3 years ago
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude
Elan Coil [88]

Answer:

The electric field is E  =  5.25 V/m

Explanation:

From the question we are told that

    The peak magnitude of the magnetic field is  B  =  17.5 nT  =  17.5 *10^{-9}\ T

Generally the peak magnitude of the electric field is mathematically represented as

         E  =  c  *  B

Where c is the speed of light with value c  =  3.0 *10^{8} \ m/s

So

       E  =  3.0 *10^{8}  *  17.5 *10^{-9}

       E  =  5.25 V/m

8 0
4 years ago
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