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Crazy boy [7]
3 years ago
5

a large cargo trucks needs to cross a bridge the truck is 30m long, and 3.2 wide, the cargo exerts a force of 54,000 N the bridg

e can withstand 450 PA of pressure. is it safe for the truck to cross the bridge?
Physics
1 answer:
otez555 [7]3 years ago
5 0

Answer:

It isn't safe for the truck to cross the bridge because the pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 Pa > 450 Pa.

Explanation:

Pressure is expressed in Force/Area.

So, for the truck, force exerted = 54000 N

Area covered = 30 × 3.2 = 96 m²

Pressure exerted by the truck = 54000/96 = 562.5 Pa

The pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 > 450, hence, it isn't safe for the truck to cross the bridge.

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What are the 2 types of ions.
Alina [70]

Answer:

Anions have more electrons than protons and so have a net negative charge. Cations have more protons than electrons and so have a net positive charge. Zwitterions are neutral and have both positive and negative charges at different locations throughout the molecule.

Explanation:

4 0
3 years ago
Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ
andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement.  So the variation of mechanical energy is equal to the work of the fictional force

    W_{fr} = ΔEm = Em_{f} -Em₀

Let's write the mechanical energy at each point

Initial

    Em₀ = Ke = ½ k x²

Final

   Em_{f} = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

    F = - k x

    x = -F / k

    x = 4400/1100

    x = - 4 m

Let's write the energy equation

    fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

   v² = (fr d + ½ kx² - mg y) 2 / m

   v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50)   2/60.0

   v² = (160 + 8800 - 1470) / 30

   v = √ (229.66)

   v = 15.8 m/s

5 0
3 years ago
Read 2 more answers
An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
What is the power of a diverging lens of focal length 0.4 m
romanna [79]

Answer:

-2.5 is the answer to your question

4 0
3 years ago
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

L_s = Latent heat of steam

s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

Heat released from water at 100 °C

Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

Heat released from water at 0 °C

Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

Total heat released is

Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s

The velocity of the ice would be 2452.79432 m/s

6 0
4 years ago
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