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denis-greek [22]

3 years ago

15

To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per

-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop 0.650×10−15 m in radius carrying 1.05×104 A . What is the field at the center of such a loop?

1 answer:

gtnhenbr [62]3 years ago

7 0

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

$B = \dfrac{\mu_0NI}{2R}$

$B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}$

B = 4.059 x 10¹⁵ T

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A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

Brut [27]

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

8 0

3 years ago

What information do you need to describe an object's location

stira [4]

longitude and latitude<span />

4 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

3 years ago

A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog

Hunter-Best [27]

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

6 0

3 years ago

In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical

mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = $\dfrac{mass\ of\ string}{length}$

linear density of the string = $\dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}$

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0

3 years ago