Physics, 03.04.2020 16:27 devnnn44

Find the terminal speed of the rod if it has mass m = 2.8 grams , length l = 21 cm , and resistance R = 0.0011 Ω . It is falling in a uniform horizontal field B = 0.052 T . Neglect the resistance of the U-shaped conductor.

Answers: 3

Show answers

Answers

Answer from: bebo14

0.253 m/s

Explanation:

As the conductor falls down, the magnetic flux throug the coil formed by the conductor and the the rest of the circuit changes, therefore an electromotive force is induced in the rod; its magnitude is given by

E=BvL

where

B = 0.052 T is the strength of the magnetic field

v is the speed at which the rod is falling

L = 21 cm = 0.21 m is the length of the rod

Due to this electromotive force, a current is also induced in the rod and the circuit, and this current is given by

$I=\frac{E}{R}$

where

$R=0.0011 \Omega$ is the resistance of the rod

So the current is

$I=\frac{BvL}{R}$ (1)

At the same time, we know that a current-carrying wire in a magnetic field experiences a force, which is given by

F_B = IBL (2)

where in this case:

I is the induced current given by eq(1)

B is the strength of the magnetic field

L is the lenght of the rod

Inserting eq(1) into (2), we find that the magnetic force on the rod is:

$F_B=\frac{BvL}{R}\cdot BL = \frac{B^2 L^2 v}{R}$

However, there is another force acting on the rod: the force of gravity, given by

F_g=mg

where

m=2.8 g = 0.0028 kg is the mass of the rod

g=9.8 m/s^2 is the acceleration due to gravity

The falling rod will reach its terminal speed when its net acceleration becomes zero; this occurs when the net force on it is zero, so when the magnetic force is balanced by the force of gravity, so when

F_B = F_g

$\frac{B^2 L^2v}{R}=mg$

And solving for v, we find the terminal speed:

$v=\frac{mgR}{B^2L^2}=\frac{(0.0028)(9.8)(0.0011)}{(0.052)^2(0.21)^2}=0.253 m/s$

Answer from: Quest

The low lands i believe

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Another question on Physics

Physics, 21.06.2019 22:00

In a wind tunnel the speed changes as the cross sectional area of the tunnel changes. if the speed in a 6' x 6' square test section is 100 mph, what was the speed upstream of the test section where the tunnel measured 20' x 20'? use conservation of mass and assume incompressible flow. conservation of mass requires that as the flow moves through a path or a duct the product of the density, velocity and cross sectional area must remain constant; i.e., that ova-constant. a model is being tested in a wind tunnel at a speed of 100 mph if the flow in the test section is at sea level standard conditions, what is the pressure at the model's stagnation point? (a) the tunnel speed is being measured by a pitot-static tube connected to a u- tube manometer. what is the reading on that manometer in inches of water? (b) at one point on the model a pressure of 2058 psf is measured. what is the local airspeed at that point?

Answers: 2

Answer

Physics, 22.06.2019 11:20

If a rock is thrown upward on the planet mars with a velocity of 12 m/s, its height (in meters) after t seconds is given by h = 12t − 1.86t2. (a) find the velocity of the rock after two seconds. m/s (b) find the velocity of the rock when t = a. 12−3.72a m/s (c) when will the rock hit the surface? (round your answer to one decimal place.) t = s (d) with what velocity will the rock hit the surface? m/s

Answers: 1

Answer

Physics, 22.06.2019 15:30

What are the necessary conditions for fusion?

Answers: 1

Answer

Physics, 22.06.2019 23:50

An alpha particle with kinetic energy 12.5 mev makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude l = p0 b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.40×10−12 m. (assume that the lead nucleus remains stationary and that it may be treated as a point charge. the atomic number of lead is 82. the alpha particle is a helium nucleus, with atomic number 2.) what is the distance of closest approach?

Answers: 2

Answer

You know the right answer?

Find the terminal speed of the rod if it has mass m = 2.8 grams , length l = 21 cm , and resistance...

Questions