How many coulombs of charge are needed to produce 29.1 mol of solid zinc?

A long, straight wire carrying a current of 2.60 A is placed along the axis of a cylinder of radius 0.500 m and a length of 4.25

shutvik [7]

<h2>Answer:</h2>

0

<h2>Explanation:</h2>

Since the current carrying wire is placed along the axis of the cylinder, according to the right hand rule, the magnetic field will be tangent to the surface of the cylinder. Therefore, there is no magnetic field through the cylinder.

Remember that the magnetic flux through a given area is the total magnetic field passing through that area. Since there is not magnetic field through the cylinder, the total magnetic flux is therefore zero (0).

7 0

3 years ago

Gaining neutrons makes an element?

Kobotan [32]

Answer:

C. Have no change in electrical charge

Explanation:

If a element gains neutron it become an Isotope. The electrical charge do not change for this, only the atomic mass changes when an element gains neutrons.

The electrical charge is affected when there is a variation in the number of electrons or protons in the element.

6 0

3 years ago

A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the

Harlamova29_29 [7]

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.

4 0

3 years ago

You have a circular loop carrying wire in a large uniform magnetic field because it's a current-carrying loop in a magnetic fiel

Mama L [17]

Answer:

The torque experienced by the new loop is 1/4 of that experienced by the old loop.

Explanation:

The torque τ = IBASinΦ

Assuming the direction of current flow through the loop is perpendicular to that of the magnetic field, SinΦ = Sin90° = 1 and

τ = IBA. With the current flowing through the loop and the magnetic field remaining the same for both cases, then

τ ∝ A. The area is equal to πD² where D is the diameter so

τ ∝ D². Let the old loop have a diameter D. Using the same length of wire, the act of doubling the loop (making 2 turns of wire) reduces the diameter by a factor of 1/2. So the new diameter is D/2

τ ∝ (D/2)² which means that τ ∝ D²/4. So the new torque is 1/4 the old torque.

5 0

3 years ago

A. When light of wavelength 250 nm is incident on a metal surface, the maximum speed of the photoelectrons is 4.0 × 105 m/s, wha

aliina [53]

Answer:

Explanation:

Part A

- Work function ( ∅ ) is the minimum energy required by the photon to knock an electron out of the metal surface. That is the portion of energy of a photon transferred to an electron so that it can escape the metal.

- We can mathematically express it:

∅ = Ep - Ek

= $\frac{h*c}{lamba} - 0.5*m_e*v^2_e\\$

Where,

Planck's constant ( h ) = 6.6261*10^-34

Speed of light ( c ) = 3*10^8 m/s

mass of an electron ( m_e ) = 9.1094*10^-31 kg

Given:-

Incident light's wavelength ( λ ) = 250*10^-9 m

The maximum speed o electron ( v_e ) = 4*10^5 m/s

Solution:-

- Plug the values into the expression derived before:

∅ = $\frac{(6.6261*10^-^3^4)*(3*10^8)}{250*10^-^9} \\\\$

∅ = $7.22257*10^-^1^9 J * \frac{6.242*10^1^8 eV}{J}$

∅ = 4.508 eV ... Answer

Part B

- 2.5% of the energy emitted by a 100-W light bulb was visible light with wavelength ( λ ) = 500*10^-9 m. In 1.0 min the amount of energy harbored by a stream of photons are:

$E_p = n_p*\frac{h*c}{lambda} = P*t*e$

Where,

Planck's constant ( h ) = 6.6261*10^-34

Speed of light ( c ) = 3*10^8 m/s

Given:-

visible light's wavelength ( λ ) = 500*10^-9 m

Power of light bulb ( P ) = 100 W

Time taken ( t ) = 1.0 min = 60 s

Portion of energy as light ( e ) = 0.025

Solution:-

- Plug the values into the expression derived before:

$n_p = \frac{(lambda)*(P)*(t)*(e)}{h*c} \\\\n_p = \frac{(500*10^-^9)*(100)*(60)*(0.025)}{(6.6261*10^-^3^4)*(3*10^8)} \\$

n_p = 3.773 * 10^20 ... Answer

Part C

- A blood vessel of radius ( r ) with length ( L ) carries blood with viscosity ( μ ). The pressure drop ( ΔP ) in the blood vessel was witnessed.

- Pressure loss ( ΔP ) in a cylindrical blood vessel is given by the Darcy's equation given below:

$\frac{dP}{p} = f*\frac{L}{D}*\frac{v^2}{2}$

Where,

ρ: Density of blood

f: Friction factor

D: Diameter of vessel

v: Average velocity

- The friction factor is a function of Reynolds number and relative roughness of blood vessel. We will assume the blood vessel to be smooth, round and the flow to be laminar ( later verified ).

- The flow rate ( Q ) in a smooth blood vessel subjected to laminar flow conditions is given by the Poiseuille's Law. The law states:

$Q = \frac{\pi*dP*r^4 }{8*u*L}$

- The velocity ( v ) in a circular tube is given by the following relation:

$v = \frac{Q}{\pi*r^2 }$

Given:-

dP ( Pressure loss ) = 2.5 Pa

radius of vessel ( r ) = 10μm = 10*10^-6 m

viscosity of blood ( μ ) = 0.0027 Pa.s

Length of vessel ( L ) = 1μm = 10^-6 m

Solution:-

- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood in the vessel:

$Q = \frac{\pi*(2.5)*(10*10^-^6)^4 }{8*(0.0027)*(10^-^6)}$

Q = 3.6361*10^-12 m^3 / s

- The corresponding velocity ( v ) of the blood flow would be:

$v = \frac{3.6361*10^-^1^2}{\pi*(10*10^-^6)^2 }$

v = 0.01157 m/s

- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood blood in the enlarged vessel ( r = 12 μm = 10*10^-6 m ) :

$Q = \frac{\pi*(2.5)*(12*10^-^6)^4 }{8*(0.0027)*(10^-^6)}$

Q = 7.54*10^-12 m^3 / s

- The corresponding velocity ( v ) of the blood flow would be:

$v = \frac{7.53982*10^-^1^2}{\pi*(12*10^-^6)^2 }$

v = 0.01666 m/s

Part D

- A radioactive isotope of Cobalt ( Co - 60 ) undegoes Beta decay ( 0.31 MeV ) and emits two gamma rays of energy ( 1.17 & 1.33 ) MeV.

- The radioactive decay for the ( Cobalt - 60 ) can be expressed in form of an equation:

$_2_7Co ( 60 ) ---> _2_8 Ni ( 60 ) + e^- + v_e^- + gamma$

- The half life ( T_1/2 ) of the Co-60 can be used to determine the decay constant ( λ ):

λ = $lambda = \frac{Ln(2)}{T_1/2}\\$

Where,

T_1/2 = 5.2 yrs = 1.68*10^8 s

Hence, the decay constant is

λ = $\frac{Ln ( 2 ) }{1.68*10^8}$ = 4.22*10^-9 s^-1

- The activity ( A ) of any radioactive isotope is function of time ( t ) defined by negative exponential distribution:

$A = A_o*e^(^-^l^a^m^b^d^a^*^t^)$

Where,

A_o: The initial activity ( Bq )

- The activity of the radioactive isotope Co-60 was A = 10 Ci after t = 30 months. The initial activity ( A_o ) can be determined:

$A_o= \frac{A}{e^(^-^l^a^m^b^d^a^*^t^)} \\\\A_o= \frac{(10Ci)*(3.7*10^1^0 Bq/Ci)}{e^(^-^4^.^2^2^*^1^0^-^9*^7^.^8^8^*^1^0^7^)} \\$

A_o = 5.016 * 10^11 Bq

- The initial number of nuclei in the sample ( N_o ) is given by:

$N_o = (A_o) / (lambda)\\\\N_o = \frac{5.016*10^1^1}{4.22*10^-^9} = 1.22*10^2^2$

- The initial mass of Co-60 used as a sample can be determined:

$m_o = M_r*N_o\\\\m_o = (59.933822u)*(1.66*10^-^2^7)*(1.22*10^2^2)\\$

m_o = 12.2 * 10^-6 kg ... Answer

- The total energy ( E ) released from the beta decay transformation:

E = E(β) + E(γ1) + E(γ2) = 0.31 + 1.17 + 1.33 = 2.81 MeV

- The rate at which the source emits energy after 30 months:

P = E*A = $( 2.81 MeV * \frac{1.6*10^-^1^3 J}{MeV} ) * ( 10 Ci * \frac{3.7*10^1^0 Bq}{Ci} )$

P = 0.166 W .. Answer

7 0

3 years ago