The correct answer is C. The level of competition is not very high in most Masters programs.
Explanation:
In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters programs".
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all directions. What is the waves intensity 25 km from the antenna
Answer:
[tex]I=2.67\times 10^{-6}\ W/m[/tex]
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m[/tex]
So, the wave intensity 25 km from the antenna is [tex]2.67\times 10^{-6}\ W/m[/tex].
Gun was fired with a muzzle velocity of 350m/s, mounted at an angle of 45’ above the ground. Neglecting air resistance, compute for the following;
*Maximum height reached
*Range of the projectile
*Total time of flight
Answer:
Maximum height, h = 3062.5m
Total time of flight, T = 49.49secs or 50secs approx.
Range, R = 12250m
Explanation:
Given data:
U = 350m/s
Angle = 45°
Assume g = 10m/s
At the greatest height, v = 0
Therefore,
V^2 = U^2 sin^2 × angle - 2×g×h
Substituting values:
0^2 = 350^2 sin^2 (45) - 2 × 10 × h
Let h = maximum height reached
Rearranging gives:
350^2 sin^2(45) = 2 x 10 x h
h = 350^2 sin^2(45)/2×10
h = 122500 x 0.5/20
h = 61250/20
h = 3062.5m
2)Total time of flight, T
T = 2U sin(angle)/g
= 2x350 sin(45)/10
= 494.9747/10
= 49.49secs or 50sec approx.
3) Range of projectile, R
R = U^2 sin2(angle)
= 350^2 sin2 (45)
= 122500 x 1/10
= 12250m
why does a hot-air ballon rise? a the volume of the air dispalced by the balloon is less than the volume of the balloon. b the weight of the air dispaced by the ballon is less than the volue of the ballon. c the wight of the balloon is less than the weight of the air displced by the balloon.
Answer:
the answer is c
Explanation:
the balloon is lighter than the air around displaced by it
The voltage across a membrane forming a cell wall is 84.0 mV and the membrane is 9.40 nm thick. What is the electric field strength (in V/m)
Answer:
8.9*10^6 V/m
Explanation:
The expression for electric field strength E is given as
[tex]E=\frac{V}{d}[/tex]
where V= voltage
d= distance of separation
Given data
[tex]voltage= 84 mV= 84*10^-^3\\distance= 9.4*10^-^9[/tex]
substituting our given data into the electric field strength formula we have
[tex]E= \frac{84*10^-^3}{ 9.4*10^-^9} \\\\E= \frac{84}{9.4} *10^-^3^-^(^-^9^)\\\\E= \frac{84}{9.4} *10^-^3^+^9\\\\E= \frac{84}{9.4} *10^6\\\\E=8.9*10^6 V/m[/tex]
The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing length will be:
A) 90 cm
B) 60 cm
C) 50 cm
D) 75 cm
Answer:
Potential gradient = 1.6 v/m
E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l
l=1.2\1.6
=3\4
3\4=0.75m
convert m into cm 0.75x10
0.75m=75 cm
ans=75 cm
Explanation:
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
Suppose Young's double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?
Answer:
The bright fringes will appear much closer together
Explanation:
Because λn = λ/n ,
And the wavelength of light in water is smaller than the wavelength of light in air. Given that the distance between bright fringes is proportional to the wavelength
4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
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A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.
Required:
Find the change in kinetic energy.
Answer:
20J
Explanation:
The computation of the change in kinetic energy is shown below:
As we know that
Work was done on system = change in potential energy + change in kinetic energy + change in thermal energy
-430J = -510J + change in kinetic energy + 100J
-430J = -410J + change in kinetic energy
So, the change in kinetic energy is 20J
We simply applied the above formula to find out the change in kinetic energy
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
1) What is the highest atomic number element a red dwarf star can produce in its core? a. Carbon b. Oxygen c. Helium d. Iron
2) What is the highest atomic number element that can be produced in the cores of the largest stars?a. Helium b. Oxygen c. Iron d. Carbon
3) If formed at the same time, a red dwarf star is likely to become a white dwarf faster than a Sun-like star would. a) True b) False
Answer:
1) c. Helium
2) Iron
3) False.
Explanation:
1. Red dwarf is the smallest and the coolest star on the sequence. These are common stars in the milky way. Red dwarfs contains metals and the elements with higher atomic number. It is found that Helium is produced in red dwarf stars.
2. Iron is the highest atomic number element that is produced in cores of largest stars. The highest mass stars can make all elements up to iron, which is the heaviest element they can produce.
3. The end of stars life is dependent on the mass they are born with. It is not necessary that all red dwarf stars will become white dwarf stars faster than sun like star.
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.
Answer:
10 kgm/s
Explanation:
Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.
From the question,
ΔM = m(v-u)...................... Equation 1
Where ΔM = change in momentum, u = initial velocity, v = final velocity.
Note: Let upward direction be negative, and downward direction be positive.
Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s
Substitute into equation 1
ΔM = 0.2(-20-30)
ΔM = 0.2(-50)
ΔM = -10 kgm/s.
The negative sign shows that the change in momentum is Upward
The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Given data:
The mass of rubber ball is, m = 0.2 kg.
The initial speed of ball is, u = 30 m/s.
The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)
The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,
[tex]p= m ( v-u)[/tex]
Solving as,
[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]
Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
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At the scene of an accident, there are skid marks which are 200 feet long showing where a car put on the brakes. The driver swears that he was obeying the speed limit. If we assume that the car braked with a constant deceleration of 16ft/sec^2, we can determine whether or not the driver is telling the truth. how fast was he going when he applied the brakes?
Answer:
He was going at 54.5mph
Explanation:
Measuring distances in feet and time (starting with the brakes being applied) in seconds,
we have a(t) = −16, s(0) = 0, and v(T) = 0, s(T) = 200,
where T is the time of braking.
So Finding antiderivatives, v(t) = −16t + A and so s(t) = −8t²+ At +B. Since s(0) = 0,
we have B = 0. Since v(T) = 0,
we have T = A /16.
Putting this into s() gives
200 = −8(A/ 16)² +A· (A /16) = A²/32.
This gives A2 = 6,400, so A = 80 = v(0). That is, the initial speed was 80 ft/s ≈ 54.5 mph.
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
If you have completely polarized light of intensity 125 W/m2, what will its intensity be after passing through a polarizing filter with its axis at an 89.5° angle to the light's polarization direction?
Answer:
When we have completely polarized light with intensity I0, and it passes through a polarizing filter with its axis at an angle θ with respect to the light's polarization direction, the new intensity of the light will be:
I = I0*cos^2(θ)
This is called the "Malus' law".
in this case, we have:
I0 = 125 W/m^2
θ = 89.5°
then:
I = (125 W/m^2)*cos^2(89.9°) = 0.00038 W/m^2
radiation transfers energy through___. a metal. b liquid. c touch. d waves.
Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.
Answer:
The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 11^o[/tex]
The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as
[tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]
Where [tex]\beta[/tex] is mathematically represented as
[tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]
[tex]\beta = 708.1 \ rad[/tex]
So
[tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The moment about the origin O of the force F is Mo
Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = [tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right][/tex]
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Question 5
A fidget spinner that is 4 inches in diameter is spinning clockwise. The spinner spins at 3000
revolutions per minute.
At t = 0, consider the point A on the outer edge of the spinner that is along the positive horizontal
axis. Let h(t) be the vertical position of A in inches. Suppose t is measured in minutes. Find a
sinusoidal function that models h(t).
h(t) =
Can someone please help me for this question?!!!!! ASAP?!!!!
Answer:
h = 4 sin (314.15 t)
Explanation:
This is a kinematics exercise, as the system is rotating at a constant speed.
w = θ / t
θ = w t
in angular motion all angles are measured in radians, which is defined
θ = s / R
we substitute
s / R = w t
s = w R t
let's reduce the magnitude to the SI system
w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s
let's calculate
s = 314.16 4 t
s = 1,256.6 t
this is the value of the arc
Let's find the function of this system, let's use trigonometry to find the projection on the x axis
cos θ = x / R
x = R cos θ
x = R cos wt
projection onto the y-axis is
sin θ = y / R
how is a uniform movement
θ = w t
y = R sin wt
In the case y = h
h = R sin wt
h = 4 sin (314.15 t)
a bear has a mass of 500kg and 100,000 J of mechanical kinetic energy. What is the speed of the bear? (velocity) Can someone please answer with the formula included? Step by step pls.
Answer:
Velocity of bear (v) = 20 m/s
Explanation:
Given:
Mass of bear (m) = 500 kg
Mechanical kinetic energy (K.E) = 100,000 J
Find:
Velocity of bear (v) = ?
Computation:
Mechanical kinetic energy (K.E) = 1/2(m)(v)²
100,000 = 1/2(500)(v)²
200,000 = 500(v)²
400 = (v)²
Velocity of bear (v) = 20 m/s
Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. Part A If a boat and its riders have a mass of 1200 kg and the boat drifts in at 1.3 m/s how much work does Sam do to stop it
Answer:
W = 1014 J = 1.014 KJ
Explanation:
As, Sam has to stop the boats in the log ride. Therefore, the work Sam needs to do, in order to stop a boat must be equal to the kinetic energy of the boat:
Work Done by Sam = Kinetic Energy of the Boat
W = K.E
W = (1/2)mv²
where,
m = mass of boat and its rider = 1200 kg
v = speed of the boat = 1.3 m/s
Therefore,
W = (1/2)(1200 kg)(1.3 m/s)²
W = 1014 J = 1.014 KJ
A stone is thrown vertically upward with a speed of 21.0 m/s . How fast is it moving when it is at a height of 11.0 m ?
Answer:
15.01 m/s
Explanation:
Applying newtons equation of motion,
v² = u²+2gs.................. Equation 1
Where v = final velocity, u = initial velocity, s = distance, g = acceleration due to gravity.
Given: u = 21 m/s, s = 11 m
Constant: g = -9.8 m/s² ( g is negative because the stone against the force of gravity).
Substitute these values into equation 1
v² = 21²-2(11)(9.8)
v² = 441-215.6
v² = 225.4
v = √225.4
v = 15.01 m/s
A system consists of two particles. The first particle has mass m1 = 6.60 kg and a velocity of (4.20i + 2.00j) m/s, and the second particle has mass m2 = 2.00 kg and a velocity of (2.00i − 3.60j) m/s. (Express your answers in vector form.)
Required:
a. What is the velocity of the center of mass of this system?
b. What is the total momentum of this system?
Answer:
a. 8.465 m/s
b.22.3659
A brass ring of diameter 10.00 cm at 20.0 ^ { \circ } \mathrm { C }20.0 ∘C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0 ^ { \circ } \mathrm { C }20.0 ∘C. Assuming the average coefficients of linear expansion are constant, What if the aluminum rod were 10.02 cm in diameter?
Answer:
b)using
DT = (LAl - LBr) / (LBr aBr - LAl aAl)
DT = (10.02-10)/(10*19x10-6 –10.02*24x10-6)
DT = -396 C°
20°C + -396 C° < -273.15 °C;
So the temperature will be -396° this is unattainable because we can’t go below absolute zero
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is
Explanation:
F = kx
mg = kx
(20 kg) (10 m/s²) = (1800 N/m) x
x = 0.11 m
A particle with a charge of 4.0 μC has a mass of 5.0 × 10 -3 kg. What electric field directed upward will exactly balance the weight of the particle?
Answer:
E = 12.25 x 10³ N/C = 12.25 KN/C
Explanation:
In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,
Electrostatic Force = Weight
E q = mg
where,
E = Electric Field = ?
m = Mass of the Charge = 5 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
If d is 10% smaller than D, how do the pressure drops experienced by the fluid across region I and region II compare?
Answer:
ΔP = - 0.262 ρ v₁²
Explanation:
This is a fluid mechanics problem, let's use the subscript for region I and the subscript for region II. Let's write Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
suppose the pipe is horizontal, therefore
y₁ = y₂
P₁ -P₂ = ½ ρ (v₁² -v₂²) 1
Now let's use the continuity equation
v₁ A₁ = v₂ A₂ 2
The area of a circle is
A = π r²
also the radius is half the diameter r = d /2
A₁ = π d₁² / 4
let's substitute in equation 2
v1 π d₁² / 4 = v2 π d₂² / 4
v₁ d₁² = v₂ d₂²
v₂ = v₁ d₁² / d₂²
let's substitute in equation 1
P₁ - P₂ = ½ ρ (v₁² - v₁² (d₁² / d₂²)² )
Now let's use that the diameter d₂ = d is 10% smaller than the larger diameter d₁ = D, we assume that it is in region 1
d₂ = D -0.1D = 0.9 D
we substitute in the previous equation
P₁ - P₂ = ½ ρ v₁² (1 - (D / 0,9D)⁴)
P₁- P₂ = ½ ρ v₁² (1 - 1 / 0,9⁴ )
P1 - P2 = ½ ρ v12 (- 0.524)
ΔP = - 0.262 ρ v₁²
in this solution we assume that the data in zone I is known
which of the following is true about scientific models?
A. models are used to simplify the study of things
B.computer models are the most reliable kind of model
C. models explain past, present and future information
D.a model is accurate if it does not change over time
Answer:
A- to simplify the study of things
Explanation:
a visual reference
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ............ cm
(b) the time interval required to complete one revolution ............ s
Answer:
(a) 3.9cm
(b) 1.66 x 10⁻⁸s
Explanation:
Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,
F = m x a --------------(i)
Where;
m = mass of the particle
a = acceleration of the mass
The centripetal acceleration is given by;
a = v² / r [v = linear velocity of particle, r = radius of circular path]
Therefore, equation (i) becomes;
F = m v²/ r --------------------(ii)
The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;
F = qvBsinθ -------------(iii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = angle between the velocity and the magnetic field
Combine equations (ii) and (iii) as follows;
m (v² / r) = qvBsinθ [divide both side by v]
m v / r = qBsinθ [make r subject of the formula]
r = (m v) / (qBsinθ) ---------(iv)
(a) From the question;
v = 1.48 x 10⁷m/s
B = 2.14mT = 2.14 x 10⁻³T
θ = 90° [since the direction of velocity is perpendicular to magnetic field]
m = mass of electron = 9.11 x 10⁻³¹kg
q = charge of electron = 1.6 x 10⁻¹⁹C
Substitute these values into equation (iv) as follows;
r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)
r = 3.9 x 10⁻²m
r = 3.9cm
Therefore, the radius of the circular path is 3.9cm
(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by
T = d / v --------------(*)
Where;
d = distance traveled in the circular path in one complete turn = 2πr
v = velocity of the motion = 1.48 x 10⁷m/s
d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]
d = 2(3.142)(3.9 x 10⁻²) = 0.245m
Substitute the values of d and v into equation (*) as follows;
T = 0.245 / 1.48 x 10⁷
T = 0.166 x 10⁻⁷s
T = 1.66 x 10⁻⁸s
Therefore, the time interval is 1.66 x 10⁻⁸s