Suppose two carts, one twice as massive as the other, fly apart when the

A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of

Veseljchak [2.6K]

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

$B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}$

Put all the values,

$I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A$

So, it carry current of 2172.46 A.

7 0

3 years ago

Which statement best compares the momentum of a 5-kilogram fish swimming at a speed of 10 meters/second and a 2-kilogram fish sw

ruslelena [56]

M = mass of the larger fish =5kg
V = velocity of the larger fish =10m/s
m = mass of the smaller fish =2kg
v = velocity of the smaller fish =10m/s
formula=
MV = mv
5kg*10m/s=2kg*10m/s
biggern mass fish has more momentum
hope this helps

6 0

3 years ago

Read 2 more answers

what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s

kotegsom [21]

Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.

7 0

3 years ago

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a

weeeeeb [17]

Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

$F \alpha \frac{1}{d^{2}}$ .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

$F' \alpha \frac{1}{9d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

$\frac{F'}{F}=\frac{d^{2}}{9d^{2}}$

$F'=\frac{F}{9}$

Thus, the force becomes one - ninth times the initial force.

6 0

3 years ago

A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that

kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0

3 years ago