The perimeter of the regular hexagon is 624 cm. If the length of one side is 7x

A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate

MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, $m=1.16\times 10^{-26}\ kg$

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

q is the charge on an electron

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}$

v = 78608.58 m/s

$v=7.86\times 10^4\ m/s$

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

$qvB=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}$

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0

3 years ago

An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the d

Oksanka [162]

Answer:

a) 7.35 x 10¹³ m/s²

b) 5.03 x 10⁻⁸ sec

c) 9.3 cm

d) 6.23 x 10⁻¹⁸ J

Explanation:

E = magnitude of electric field = 418 N/C

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of the electron = 9.1 x 10⁻³¹ kg

a)

acceleration of the electron is given as

$a = \frac{qE}{m}$

$a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}$

a = 7.35 x 10¹³ m/s²

b)

v = final velocity of the electron = 3.70 x 10⁶ m/s

v₀ = initial velocity of the electron = 0 m/s

t = time taken

Using the equation

v = v₀ + at

3.70 x 10⁶ = 0 + (7.35 x 10¹³) t

t = 5.03 x 10⁻⁸ sec

c)

d = distance traveled by the electron

using the equation

d = v₀ t + (0.5) at²

d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²

d = 0.093 m

d = 9.3 cm

d)

Kinetic energy of the electron is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²

KE = 6.23 x 10⁻¹⁸ J

4 0

4 years ago

A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

MariettaO [177]

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R , R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I α

= I a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

5 0

3 years ago

Shearing of the wool is done with special instruments called_____

Kobotan [32]

Answer:

The machine used is called a squaring shear, power shear, or guillotine.

Explanation:

5 0

3 years ago

Read 2 more answers

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

Object 3 should be placed closer to object 1.

Object 3 should be placed on a halfway between object 2 and object 1.

Object 3 should be placed closer to object 2.

Solution

I think that Object 3 should be placed closer to object 2.

6 0

3 years ago

The perimeter of the regular hexagon is 624 cm. If the length of one side is 7x - 1,