All categories

2 years ago

Compare and contrast a tanker truck carrying a liquid load and a truck carrying solid objects

Physics

1 answer:

givi [52]2 years ago

4 0

Answer:

Explanation:

When carrying a liquid load the tanker will have pressure on a larger area of the container as compared to a tanker truck carrying a solid load.

Since the liquid has a property to flow therefore the molecules of the liquid load throughout the bulk will fluctuate more on any movement or vibration than the solid load.

The liquid load will require a leak proof container whereas the solid load container has no such restrictions.

The solid load will exert the pressure majorly on the bottom surface of the tanker truck whereas the pressure is distributed over the area of contact in case of the liquid load.

Solid load can be bound by the fixtures whereas the liquid load requires compartments in the tanker to minimize the fluctuations on acceleration and deceleration.

You might be interested in

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

6 0

2 years ago

Determine the torque applied to the shaft of a car that transmits 225 hp

Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

$T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }$

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

$T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu$

8 0

2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

7 0

2 years ago

Read 2 more answers

Neha and Suhani are playing with two identical pendulums. They leave the bob from a certain position and wait for it to return t

Anika [276]

Answer:

The correct option is C

Explanation:

The pendulum bob would return at the same time because the initial angle a pendulum bob is dropped does not affect it's period (the time it takes for the pendulum to move back and forth), however the one with a larger angle move faster but would eventually arrive at the same "starting point" due to varying displacements made.

6 0

2 years ago

An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and

serg [7]

<h2>The flux through the infinite charged wire along the central axis of a cylindrical surface of radius r and length l is ∅E = E x 2πrl </h2>

Explanation:

let us consider a thin infinitely long straight wire having a uniform charge density λ Cm⁻¹.To determine the field at a distance r from the line charge , we have cylindrical gaussian surface of radius r, length l,and with its axis along the line charge. it has curved surface S₁ , and flat circular ends S₂ and S₃. Obviously, dS₁//E, dS₂ ⊥E , and dS₃ ⊥ E , so, only the curved surface contributes towards the total flux.

∅E = ∫ E.dS = ∫E.dS₁ +∫E.dS₂ +∫E.dS₃

= ∫EdS₁ cos0⁰ +∫EdS₂ cos 90⁰ +∫Eds₃ cos 90⁰

= E∫ds₁₁ +0+0

= E x area of curved surface

∅E = E x 2πrl

3 0

2 years ago