g A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75 C/s . Part A How many electrons are delivered to the

Answers

Answer 1

Answer:

Approximately 5 x 10^18 electrons are delivered to the smart phone charger.

Explanation:

The electric current in a circuit is the flow of charges through a circuit with time.

The charges through the circuit are due to the electrons that flow through the circuit.

An individual electrons has a charge of -1.60 x 10^-19 C on it.

If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.

==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.


Related Questions

A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?

Answers

Answer:

counterclockwise

Explanation:

given data

area = 25 cm²

solution

We know that a changing magnetic field induces the current and induced emf is express as

[tex]\epsilon = -N \frac{d \phi }{dt}[/tex]     ..................................1

and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field

so when magnetic field decrease and point coming out of the paper.

so induced current in the loop will be counterclockwise

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled

Answers

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

16. If one body is positively charged and another body is negatively charged, free electrons tend to
O A. move from the negatively charged body to the positively charged body
O B. remain in the positively charged body
OC. move from the positively charged body to the negatively charged body
O D. remain in the negatively charged body

Answers

Answer:

Hey there!

The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body

Let me know if this helps :)

a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?​

Answers

Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .

Explanation:

The average speed of the car is 31,680 meters per hour.

To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.

Given:

Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours

Distance traveled (d) = 5,280 meters

Average Speed (v) = Distance (d) / Time (t)

Average Speed (v) = 5280 meters / (1/6) hours

To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:

Average Speed (v) = 5280 meters × (6/1) hours

Average Speed (v) = 31,680 meters per hour

Hence, the average speed of the car is 31,680 meters per hour.

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In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.

Answers

Answer:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:  speed of light(c)

Explanation:

Generally the ratio of the E(electric field ) and  the B(magnetic field ) is  equal to the speed of the electromagnetic wave i.e the speed of  light (c) the value is

    [tex]c = 3.0 *10^{8} \ m/s[/tex]

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T

Answers

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

[tex]F = \frac{mv^2}{r}[/tex]

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire

Answers

Answer:

The direction of the force is towards the East.

Explanation:

Using the right hand rule, the force on the current carrying conductor is east.

In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.

In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.

A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.

Answers

Answer:

82.2 mL

Explanation:

The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;

C1V1=C2V2

Where;

C1= concentration of stock solution

V1= volume of stock solution

C2= concentration of dilute solution

V2= volume of dilute solution

V2= C1V1/C2

V2= 1.48 × 55.6/ 1.0

V2= 82.2 mL

A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?

Answers

Answer:

The amplitude [tex]A(5) = 1 \ m[/tex]

Explanation:

From the question we are told that

     The  spring constant is  [tex]k = 100 \ N/m[/tex]

      The  damping constant is  [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]

       The mass is  [tex]m = 0.050 \ kg[/tex]

       The  maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           [tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]

substituting values

         [tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]

         [tex]A(5) = 1 \ m[/tex]

       

Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact

Answers

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.

Required:
How far away did the impact occur?

Answers

Answer:

The distance is [tex]d = 193.6 \ m[/tex]

Explanation:

From the question we are told that

   The time interval between the sounds is  k[tex]t_1 = k + t_2[/tex] =  0.50 s

    The  speed of sound in air is  [tex]v_s = 343 \ m/s[/tex]

    The  speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]

 

Generally the distance where the collision occurred is  mathematically represented as

          [tex]d = v * t[/tex]

Now from the question we see that d is the same for both sound waves

 So

        [tex]v_c t = v_s * t_1[/tex]

Now  

So [tex]t_1 = k + t[/tex]

      [tex]v_c t = v_s * (t+ k)[/tex]

=>     [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]3000 t = 343* (t+ 0.50)[/tex]

=>    [tex]t = 0.0645 \ s[/tex]

So

     [tex]d = 3000 * 0.0645[/tex]

     [tex]d = 193.6 \ m[/tex]

       

     

Convert 7,348 grams to kilograms

Answers

0.00735 kilograms is your answer. I’m sorry If I didn’t explain it right :(,

A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?

Answers

Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?

Answers

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

What is pressure?

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.

Answers

volume of balloon

= 4/3 T R3

= 4/3 x 3.14 x 6.953

= 1405.47 m3

uplift force

= volume of balloon x density of air x 9.8

= = 1405.47 x 1.29 x 9.8

= 1813.05 x 9.8 N

weight of helium gas

= volume of balloon x density of helium x

9.8

= 1405.47 x .179 x 9.8

= 251.58 x 9.8 N

Weight of other mass = 930 x 9.8 N Total weight acting downwards

= 251.58 x 9.8 +930 x 9.8

= 1181.58 x 9.8 N

If W be extra weight the uplift can balance

1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8

1181.58+W=1813.05

W= 631.47 kg

Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom

Answers

Answer:

They both have the same total K.E at the bottom

Explanation:

This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved

So

K1 + U1 = K2 + U2

But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference

thus

K1 = 0 , U2 = 0

ΔK = ΔU = m g. h

But if the two inclines have the same height, and both hoops have the same mass m,

So difference in kinetic energy, must be the same for both hoops.

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?

Answers

Answer:

A) 0.0267 T

B) 0.0263 T

Explanation:

Given that

The number of turns, N = 400

Radius of the wire, r = 6 cm = 0.06 m

Current in the wire, I = 0.25 A

Relative permeability, K(m) = 80

See the attached picture for the calculation

An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?

Answers

Answer:

The  value is  [tex]N = 3.708*10^{7} \ \ atoms[/tex]

Explanation:

From the question we are told that

    The pressure is  [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]

    The  temperature is  [tex]T = 26 + 273 = 299 \ K[/tex]

     The volume is  1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]

Generally according to the ideal gas law we have that

      [tex]PV = NkT[/tex]

here  k is the Boltzmann constant with a value  [tex]k = 1.38 *10^{-23} \ J/K[/tex]

  =>  [tex]N = \frac{PV}{ k T}[/tex]

=>     [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]

=>    [tex]N = 3.708*10^{7} \ \ atoms[/tex]

     

what is defect of vision​

Answers

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar objects in their countries

Answers

Answer: no u

Explanation: no u

If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength

Answers

Answer:

0.09 x10^-10m

Explanation:

Using wavelength=( 12.27 A)/√V

= 12.27 x 10^-10/ √1.6x10^2

= 0.09x10^-10m

A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200

Answers

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!                              

The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.

Answers

Answer:

covalent

Explanation:

covalent bonds share electrons

Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?

Answers

Answer:

The final temperature is 61.65 °C

Explanation:

mass of copper pot [tex]m_{c}[/tex] = 2 kg

temperature of copper pot [tex]T_{c}[/tex] = 20 °C  (the pot will be in thermal equilibrium with the room)

specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C

The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J

mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg

temperature of boiling water [tex]T_{w}[/tex] = 100 °C

specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C

The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J

The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J

This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation

[tex]H_{T}[/tex] =   [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]

where [tex]T_{f}[/tex] is the final temperature of the water and the copper

substituting values, we have

99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x

99040 = 770[tex]T_{f}[/tex] + 836.4

99040 = 1606.4[tex]T_{f}[/tex]

[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC

Answers

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both

Answers

Answer:

a. Moon around Earth.

Explanation:

Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.

The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.

The object that should exhibit the longest orbital period is option a. Moon around Earth.

What is Charon's orbit?

Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.

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Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.

Answers

Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer

Explanation:

Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.

What is friction?

Friction is the force that prevents one hard material from scooting or rolling over the other.

Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.

We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.

Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.

In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.

Thus, the correct option is B.

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g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.

Answers

Answer:

Time, [tex]t=1.07\times 10^{-23}\ s[/tex]

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]

So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]

A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.

a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)

1. Yes
2. No

Answers

.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

ANS=1.29*10^-12 N

2. Using Amax=Fmax/ m

Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)

ANS=1.93*10^15 m/s^2*

3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton

How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?

Answers

Answer:

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

Explanation:

The decay of radioisotopes are represented by the following ordinary differential equation:

[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]m[/tex] - Mass of the radioisotope, measured in grams.

The solution of this expression is:

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.

The ratio of current mass to initial mass is:

[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]

The time constant is now calculated in terms of half-life:

[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.

Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:

[tex]\tau = \frac{88\,s}{\ln 2}[/tex]

[tex]\tau \approx 126.957\,s[/tex]

Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:

[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]

[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]

[tex]t \approx 352\,s[/tex]

352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.

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