if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then:

$I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}$

$I = 1370.05A$

The current is too high to be transported which would make the case not feasible.

8 0

3 years ago

If you go to the beach and get a cup of water out of the ocean, does the water in the cup have the same average kinetic energy a

DiKsa [7]

Yes, the water in the cup have the same average kinetic energy as the ocean, because I focus on the word 'average'. However the cup of water have less INTERNAL energy than the ocean.
hope it helps☺☺

3 0

3 years ago

Please help this is really important thanks

Juliette [100K]

1. Giga is the largest
2. Stem-and-leaf

6 0

3 years ago

Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min

marin [14]

Mary walks:
d 1 = 80 m, d 2 = 125 m, d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t
v = ( 80 m + 125 m + 45 m ) / 600 s
v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second

5 0

2 years ago

2. Using Graph 2, calculate the net force experienced by the particle between 4 and 6 seconds. The

Trava [24]

Using Newtons Second Law:

F = m×a

F = (0.25 kg)(-2 m/s²)

F = -0.5 N

<h2>The correct option is C</h2>

3 0

2 years ago

if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.​

if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.