What happens when an object with a lower density is placed in a container with an

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

OLEGan [10]

Answer:

$Relative\ Velocity = 105m/s$

Explanation:

Given

$V_A = 45m/s$

$V_B = 60m/s$

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a $sum\ of\ velocities\ of$ both trains:

This is shown below:

$Relative\ Velocity = V_A + V_B$

$Relative\ Velocity = 45m/s + 60m/s$

$Relative\ Velocity = 105m/s$

4 0

3 years ago

The ramp on the back of a moving van is in example of what type <br> simple machine

abruzzese [7]

Inclined Plane.Because an inclined plane is a flat,sloped surface. And a ramp is a perfect example of an inclined plane.

4 0

3 years ago

Read 2 more answers

List and explain briefly similarities and differences between the electric force between two charges and the gravitational force

bazaltina [42]

Answer:

Please see below as the answers are self-explanatory.

Explanation:

Similarities

1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)

2) Both type of forces obey Newton's 3rd law.

3) Both are proportional to the product of the property that is affected by the force (charges and masses)

4) Both obey an inverse - square law (consequence of our universe being three-dimensional)

Differences

1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.

2) It is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.

6 0

3 years ago

A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie

Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $$\theta$

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $$\theta$

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $$\theta$

U = Iπ $r^{2}$ B cos $$\theta$

For $$\theta$ = 0° →

U(Max) = MB cos(0) = MB = Iπ $r^{2}$ B = 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

375 π x $10^{-7}$ .

For $$\theta$ = 90° →

U = MB cos (90) = 0

For $$\theta$ = 180° →

U(Min) = MB cos(0) = - MB = - Iπ $r^{2}$ B = - 5 × π × $( 0.05 ) ^{2}$ × 3 × $10^{-3}$ =

- 375 π x $10^{-7}$ .

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ U

0° 375 π x $10^{-7}$

90° 0

180° - 375 π x $10^{-7}$

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

#SPJ4

3 0

2 years ago

g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

3 years ago