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schepotkina [342]
3 years ago
6

Abridgeis36.5mlongandweighs2.56x105N.Atruckweighing 5.25x104Nis10.2 m from one end of the bridge. Calculate the upward force tha

t must be exerted by each pier to support thisweight
Physics
1 answer:
enot [183]3 years ago
3 0

Answer:

Explanation:

Bridge length A B = 36.5 m

weight W = 2.56 x 10⁵ N acting at middle point that is at 18.25 m from either end.

weight of truck W₁ = 5.25 x 10⁴ N  standing at 10.2 m from A .

Let force at A be F₁ and at B be F₂ .

Balancing total upward and downward force

F₁ + F₂ = ( 2.56 + .525 ) x 10⁵ = 3.085 x 10⁵ N

Taking torque of all the forces , for rotational balancing

2.56 x 10⁵ x 18.25 + .525 x 10⁵ x 10.2 = F₂ x 36.5

(46.72 + 5.355 )x 10⁵ = F₂ x 36.5

F₂ = 1.42 x 10⁵ N

F₁ = 1.665 x 10⁵ N

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If a car was moving at a speed of 25mph for 6 hours, how far would the car travel?​
Anestetic [448]

Answer:

150 miles

Explanation:

25mph x 6hrs = 150miles

7 0
3 years ago
A person hums into the top of a well (tube open at only one end) and finds that standing waves are established at frequencies of
S_A_V [24]

Answer:

Explanation:

The resonance in a tube that is open at one end and closed at the other, we can find it because in the closed part you have a node and in the open part you have a belly, so for the fundamental frequency you have ¼ Lam, the different resonances are:

Fundamental              λ = 4L

3 harmonica               λ  = 4L / 3

5 Harmonica               λ = 4L / 5

General harmonica     λ = 4L / (2n-1)                n = 1, 2, 3

Let's apply this equation to our case

The speed of sound is given by

          v =  λ  f

          λ = v / f

Let's look for wavelengths

         λ₁ = 40.6 / 343 = 0.1184 m

         λ₂ = 67.7 / 343 = 0.1974 m

         λ₃ = 94.7 / 343 = 0.2761 m

Since the wavelengths are close we can assume that it corresponds to consecutive integers, where for the first one it corresponds to the integer n

          λ ₁ = 4L / (2n-1)

          λ₂ = 4L / (2 (n + 1) -1) = 4L / (2n +1)

          λ₃ = 4L / (2 (n + 2) -1) = 4L / (2n + 3)

Let's clear in the first and second equations

          2n-1 = 4L / λ₁

          2n +1 = 4L / λ₂

Let's solve the system of equations

         4L / λ₁ + 1 = 4L / λ₂ -1

         4L / λ₂ – 4L / λ₁ = 2

         2L (1 / λ₂ - 1 / λ₁) = 1

         1 / L = 2 (1 / λ₂ -1 / λ₁)

         1 / L = 2 (1 / 0.1974 - 1 / 0.1184)

         1 / L = 2 (5,066 - 8,446) = -6.76

         L = 0.1479 m

3 0
3 years ago
A 2-kg box is pushed to the right by a force of 4 N for a distance of 32 m. It has an initial velocity of 4 m/s to the right. NO
rewona [7]

Answer: a) 8 Kg m/s b) 16 Kg m/s c) 24 Kg m/s d) 16 J e) 128 J f) 144 J

              g) 4 s

Explanation:

a) As momentum by definition is the product of mass times the velocity (is a vector quantity), we can write in this case the following:

pi = m. v₀ = 2 Kg . 4 m/s = 8 Kg. m/s

b) In order to get the change in momentum, we need to get first the final speed of the object.

As we have the total distance travelled, and we could find the acceleration, we could use a kinematic equation to solve the question, but later we will need the kinetic energy, it would be better to apply the work-energy theorem, and calculate ΔK as the work done by external force F, as follows:

ΔK = F . d = 1/2 m (vf² - v₀²)

As we know F, d, m, and v₀, we can solve the equation above for vf:

vf = 12 m/s

So, we can compute the final momentum as follows:

pf = m. vf = 2 Kg. 12 m/s = 24 Kg. m/s

Finally, we can find the change in momentum, as the difference between the final momentum and the initial one, calculated in a):

Δp = pf - pi = 24 Kg. m/s - 8 Kg. m/s = 16 Kg. m/s

c) As we have already found, final momentum is as follows:

pf = m . vf = 2 Kg. 12 m/s = 24 Kg. m/s

d) By definition the initial kinetic energy of the box is as follows:

Ki = 1/2 m v₀² = 1/2. 2 Kg .4² m²/s² = 16 J

e) We can find the change in the kinetic energy taking directly the difference between the final and initial ones, as follows:

ΔK = Kf - Ki = 1/2. 2 Kg (12² - 4²) m²/s² = 128 J

f) From above, we have Kf = 1/2 m. vf² = 1/2 . 2 Kg. 12² m²/s² = 144 J

g) As we know the magnitude of F, and the value of m, we can find the acceleration (assumed constant) , applying Newton's Second Law, as follows:

Fext = m .a ⇒ a = F/m = 4 N / 2 Kg = 2 m/s²

Appying the definition of acceleration, we can solve for t, as follows:

t = (vf-v₀) / a = (12 m/s - 4 m/s) / 2 m/s² = 4 s

6 0
3 years ago
If a 1.0-kg-mass block is on the left cap, how much total mass must be placed on the right cap so that the caps equilibrate at e
Dennis_Churaev [7]

<span>The total mass that should be placed in the right cap so that the caps equilibrate at equal height is also 1 kg. if equilibrium should be maintained the force in each side should cancel out, so to balance a 1kg mass, a 1 kg mass should also be place on the opposite direction</span>

3 0
3 years ago
A pendulum has a mass of 1.5 kg and starts at a height of 0.4 m. If it is released from rest, how fast is it going when it reach
ella [17]

Answer:

A. 2.8 m/s

Explanation:

Suppose that at the height of 0 m, the path of the pendulum is lowest.

If we use law of conservation of energy, the pendulum will have zero kinetic energy or K.E when it is at highest point, because K.E happens during movement of object and at the highest point all the energy will be P.E

                                                    P.E= mgh

Similarly, when the pendulum reaches at the lowest point, the height becomes zero and the P.E also becomes zero. Now all the energy will be K.E

                                               K.E= 1/2 m v^2

In question, we are asked about the speed as the pendulum  it reaches the lowest point of its path. Like we mentioned P.E will be zero at lowest point  because of zero height. And also we will use law of conservation of energy because no energy has been lost from system.

                                                  K.E=     P.E

                                       1/2 m v^2  =   mgh

Taking sq.root at both sides

                                                  v= Under root 2 gh

                                                   v=Under root 2x 9.8 m/s x0.4 m

                                                   v=Under root 7.84

                                                    v=2.8 m/sec

Hope it helps!

3 0
3 years ago
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