Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
We are given that A city of Punjab has 15 percent chance of wet weather on any given day.
So, Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15 =0.85
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=0.85
r=3
Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]
[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]
Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]
Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Calculate the electromotive force produced by each of the battery combinations shown in the figure, if the emf of each is 1.5 V.
Answer:
A) 1.5 V
B) 4.5 V
Explanation:
A) Batteries in parallel have the same voltage as an individual battery.
V = 1.5 V
B) Batteries in series have a voltage equal to the sum of the individual batteries.
V = 1.5 V + 1.5 V + 1.5 V
V = 4.5 V
Tech A says parallel circuits are like links in a chain. Tech B says total current in a parallel circuit equals the sum of the current flowing in each branch of the circuit. Who is correct?
Answer: Only Tech B is correct.
Explanation:
First, tech A is wrong.
The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.
Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.
A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?
Answer:
9.88 milivolt
Explanation:
Given: diameter d = 5.2 cm
magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T
t = 0.29 sec.
we know emf = - dΦ/dt
and flux Φ = BA
A= area
therefore emf ε = -A(B_2-B_1)/Δt
[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]
A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck's kinetic energy
Answer:
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
Explanation:
Given:
Mass of truck(m) = 2,100 kg
Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s
Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s
Find:
Change in kinetic energy (ΔKE)
Computation:
Change in kinetic energy (ΔKE) = 1/2(m)[v2² - v1²]
Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² - 10.56²]
Change in kinetic energy (ΔKE) = 1,050[233.4784 - 111.5136]
Change in kinetic energy (ΔKE) = 1,050[121.9648]
Change in kinetic energy (ΔKE) = 128063.04
Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J
The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)
Answer:
Explanation:
Let f₀ be the frequency .
energy of photons having frequency of f₀
= hf₀ where h is plank's constant
for electron to get ejected , work function should be equal to energy of photon
hf₀ = 3.55
4.1357 x 10⁻¹⁵ x f₀ = 3.55
f₀ = 8.58 x 10¹⁴ Hz .
An inductor is hooked up to an AC voltage source. The voltage source has EMF V0 and frequency f. The current amplitude in the inductor is I0.
Part A
What is the reactance XL of the inductor?
Express your answer in terms of V0 and I0.
Part B
What is the inductance L of the inductor?
Express your answer in terms of V0, f, and I0.
Answer:
a. The reactance of the inductor is XL = V₀/I₀
b. The inductance of the inductor is L = V₀/2πfI₀
Explanation:
PART A
Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,
XL = V₀/I₀
So, the reactance of the inductor is XL = V₀/I₀
PART B
The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor
Since XL = V₀/I₀ = 2πfL
V₀/I₀ = 2πfL
L = V₀/2πfI₀
So the inductance of the inductor is L = V₀/2πfI₀
A) The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex]
B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
A) Expressing the Reactance of the inductor
Voltage across the Inductor = V₀ = I₀XL ---- ( 1 )
Where : V₀ = emf voltage , I₀ = current
from equation ( 1 )
∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]
B ) Expressing the Inductance of the Inductor
Inductance of an inductor is expressed as : XL = 2πfL
from part A
XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL
∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀
L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
Hence we can conclude that The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex] and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex] .
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From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:
a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down
Answer:
c. 1.11 m/s down
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Assuming the balloon and projectile are originally at rest:
(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)
0 kg m/s = (90 kg) v + 100 kg m/s
v = -1.11 m/s
In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.
a. True
b. False
Answer:
True or False
Explanation:
Because.....
easy 50% chance you are right
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?
Answer:
The velocity is [tex]v_h = 19.2 \ m/s[/tex]
Explanation:
From the question we are told that
The speed of the roller coaster at ground level is [tex]v = 26 \ m/s[/tex]
Generally we can define the roller coaster speed at ground level using the an equation of motion as
[tex]v^2 = u^2 + 2 g s[/tex]
u is zero given that the roller coaster started from rest
So
[tex]26^2 = 0 + 2 * g * s[/tex]
So
[tex]s = \frac{26^2}{ 2 * g }[/tex]
=> [tex]s = 37.6 \ m[/tex]
Now the displacement half way is mathematically represented as
[tex]s_{h} = \frac{37.6}{2}[/tex]
[tex]s_{h} = 18.8 \ m[/tex]
So
[tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]
Where [tex]v_h[/tex] is the velocity at the half way point
=> [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]
=> [tex]v_h = 19.2 \ m/s[/tex]
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?
Answer:
c. 70 Ω
Explanation:
The R and R resistors are in parallel. The 2R and 2R resistors are in parallel. The 4R and 4R resistors are in parallel. Each parallel combination is in series with each other. Therefore, the equivalent resistance is:
Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)
Req = R/2 + 2R/2 + 4R/2
Req = 3.5R
Req = 70Ω
A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizontal. If the initial speed of the stream is 40 m/s the height that the stream of water will strike the building is
Answer:
We can think the water stream as a solid object that is fired.
The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)
The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).
The initial speed of the stream is 40m/s.
First, using the fact that:
x = R*cos(θ)
y = R*sin(θ)
in this case R = 40m/s and θ = 30°
We can use the above relation to find the components of the velocity:
Vx = 40m/s*cos(30°) = 34.64m/s
Vy = 20m/s.
First step:
We want to find the time needed to the stream to hit the buildin.
The horizontal speed is 34.64m/s and the distance to the wall is 50m
So we want that:
34.64m/s*t = 50m
t = 50m/(34.64m/s) = 1.44 seconds.
Now we need to calculate the height of the stream at t = 1.44s
Second step:
The only force acting on the water is the gravitational one, so the acceleration of the stream is:
a(t) = -g.
g = -9.8m/s^2
For the speed, we integrate over time and we get:
v(t) = -g*t + v0
where v0 is the initial speed: v0 = 20m/s.
The velocity equation is:
v(t) = -g*t + 20m/s.
For the position, we integrate again over time:
p(t) = -(1/2)*g*t^2 + 20m/s*t + p0
p0 is the initial height of the stream, this data is not known.
Now, the height at the time t = 1.44s is
p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po
= 16.57m + p0
So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.
An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.28 diopters, and the other lens has a power of 8.50 diopters. What is the magnifying power of the telescope he can construct with these lenses
An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
I = 1.23 A
Explanation:
In an RL circuit current passing is described by
I = E / R (1 - [tex]e^{-Rt/L}[/tex])
Let's reduce the magnitudes to the SI system
L = 35 mH = 35 10⁻³ H
t = 5.0 ms = 5.0 10⁻³ s
let's calculate
I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))
I = 1.5 (1- [tex]e^{-1.715}[/tex])
I = 1.23 A
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
Answer:
a
[tex]m = 0.169 \ kg[/tex]
b
[tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 14 \ N/m[/tex]
The maximum extension of the spring is [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
The number of oscillation is [tex]n = 30[/tex]
The time taken is [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
[tex]T = \frac{t}{n}[/tex]
substituting values
[tex]T = \frac{20}{30 }[/tex]
[tex]T = 0.667 \ s[/tex]
Thus
[tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
[tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]m =\frac{k }{w^2}[/tex]
substituting values
[tex]m =\frac{ 15 }{(9.421)^2}[/tex]
[tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is mathematically represented as
[tex]v = - Awsin (wt)[/tex]
The velocity is maximum when [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
[tex]v_{max} = - A* w[/tex]
=> [tex]|v_{max} |= A* w[/tex]
=> [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> [tex]|v_{max} |= 0.5653 \ m/s[/tex]
There is a hydraulic system that by means of a 5 cm diameter plunger to which a 5 N force is applied and that force is transmitted by means of a fluid to a 1 meter diameter plunger. Determine how much force can be lifted by the 1 m diameter plunger,
1) - 234 N
2) - 800 N
3) - 636 N
4) - 600 N
Explanation:
Pressure is the same for both plungers.
P = P
F / A = F / A
F / (¼ π d²) = F / (¼ π d²)
F / d² = F / d²
5 N / (0.05 m)² = F / (1 m)²
F = 2000 N
None of the options are correct.
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until friction causes it to settle back to the equilibrium position. Ignoring friction, what maximum height will it reach above the surface
Answer:
y = 20.99 V / A
there is no friction y = 20.99 h
Explanation:
Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy
when the block of ice is submerged it is subjected to two forces its weight hydrostatic thrust
F_net= ∑F = B-W
the expression stop pushing is
B = ρ_water g V_ice
where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice
We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³
W = ρ-ice g V
F_net = (ρ_water - ρ_ ice) g V
this is the net force directed upwards, we can find the potential energy with the expression
F = -dU / dy
ΔU = - ∫ F dy
ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy
ΔU = - (ρ_water - ρ_ ice) g A y² / 2
we evaluate between the limits y = 0, U = 0, that is, the potential energy is zero at the surface
U_ice = (ρ_water - ρ_ ice) g A y² / 2
now we can use the conservation of mechanical energy
starting point. Ice depth point
Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2
final point. Highest point of the block
[tex]Em_{f}[/tex] = U = m g y
as there is no friction, energy is conserved
Em₀ = Em_{f}
(ρ_water - ρ_ ice) g A y² / 2 = mg y
let's write the weight of the block as a function of its density
ρ_ice = m / V
m = ρ_ice V
we substitute
(ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y
y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A
let's substitute the values
y = 0.913 / (1 - 0.913) 2 V / A
y = 20.99 V / A
This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction
V / A = h
where h is the height of the block
y = 20.99 h
A ferry boat sails east across a lake at 10 km/h. A woman is walking east on
the boat at 1.5 km/h. What is her speed relative to the boat?
A. 8.5 km/h west
B. 8.5 km/h east
C. 1.5 km/h east
O D. 1.5 km/h west
Answer:
B
8.5 km/h east
Explanation:
Relative velocity= Va -Vb
=10-1.5
=8.5 km/h east
The concept relative speed is used when two or more bodies moving with some speed are considered. The relative speed of woman to the boat is 8.5 km/h east. The correct option is B.
What is relative speed?The relative speed of two bodies is defined as the sum of their speeds if they are moving in the opposite direction and it is the difference of their speeds if they are moving in the same direction.
The speed of the moving body with respect to the stationary body is known as the relative speed. The term relative means in comparison to. The relative speed is a scalar quantity.
Here both the boat and women are travelling in the same direction. So the relative speed is given as:
Relative speed = 10 - 1.5 = 8.5 km / h
Therefore the relative speed is 8.5 km/h east.
Thus the correct option is B.
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A laser emits photons having an energy of 3.74 × 10–19 J. What color would be expected for the light emitted by this laser? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J ⋅ s)
Answer:
The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Explanation:
Given;
energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 x 10⁻³⁴ J.s
The wavelength of the emitted light will be calculated by applying energy of photons;
[tex]E = hf[/tex]
where;
E is the energy emitted light
h is Planck's constant
f is frequency of the emitted photon
But f = c / λ
where;
λ is the wavelength of the emitted photons
[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]
λ ≅ 532 nm
the wavelength of the emitted photons is 532 nm.
Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the cart is dragged horizontally along the floor with a constant speed of 7 km/h, calculate how much work the donkey does in 35 minutes.
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth
2. Give a specific example of a system with the energy transformation shown.
a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Answer:
1) a block going down a slope
2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK
Explanation:
In this exercise you are asked to give an example of various types of systems
1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.
2)
a) rolling a ball uphill
In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact
W = ΔU + ΔK + ΔE
b) in this system work is transformed into internal energy
W = ΔE
c) There is no friction here, therefore the work is transformed into kinetic energy
W = ΔK
d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy
ΔU = ΔK
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass.
Required:
a. First calculate the moment of inertia (in kg-m^2) when the skater has their arms pulled inward by assuming they are cylinder of radius 0.11 m.
b. Now calculate the moment of inertia of the skater (in kg-m^2) with their arms extended by assuming that each arm is 5% of the mass of their body. Assume the body is a cylinder of the same size, and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
a) The moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m².
b) If the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
Given the data in the question;
Mass of skater; [tex]M = 56.5kg[/tex]
a)
When the skater has his arms pulled inward by assuming they are cylinder of radius; [tex]R = 0.11 m[/tex]
Moment of inertia; [tex]I = \ ?[/tex]
From Parallel axis theorem; Moment of Inertia for a cylindrical body is expressed:
[tex]I = \frac{1}{2}MR^2[/tex]
Where M is the mass and R is the radius
We substitute our given values into the equation
[tex]I = \frac{1}{2}\ *\ 56.5kg\ *\ (0.11m)^2\\\\I = \frac{1}{2}\ *\ 56.5kg\ *\ 0.0121m^2\\\\I = 0.3418kg.m^2[/tex]
Therefore, the moment of inertia as the skater pulled his/her arm inward by assuming he/she is a cylinder is 0.3418kg-m²
b)
With the skater's arms extended by assuming that each arm is 5% of the mass of their body
Mass of each arm; [tex]M_a = \frac{5}{100} * M = \frac{5}{100} * 56.5kg = 2.825kg[/tex]
Remaining mass; [tex]M_b = M - 2M_a = 56.5kg - 2(2.825kg) = 50.85kg[/tex]
Assume the body is a cylinder of the same size and the arms are 0.875 m long rods extending straight out from their body being rotated at the ends.
Length of arm; [tex]L = 0.875 m[/tex]
From Parallel axis theorem; Moment of Inertia about vertical axis is expressed as:
[tex]I = \frac{1}{2}M_bR^2 + \frac{2}{3}M_aL^2[/tex]
We substitute in our values
[tex]I = \frac{1}{2}*50.85kg*(0.11m)^2 + \frac{2}{3}*2.825kg*(0.875m)^2\\\\I = [\frac{1}{2}*50.85kg * 0.0121m^2] + [\frac{2}{3}*2.825kg*0.765625m^2]\\\\I = 0.3076kg.m^2 + 1.4419kg.m^2\\\\I = 1.7495kg.m^2[/tex]
Therefore, if the body of the skater is assumed to be a cylinder of the same size, and the arms are rods extending straight out from his/her body being rotated at the ends, the moment of inertia is 1.7495kg-m².
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A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60
Answer:
-0.73mA
Explanation:
Using amphere's Law
ε =−dΦB/ dt
=−(2.6T)·(7.30·10−4 m2)/ 1.00 s
=−1.9 mV
Using ohms law
ε=V =IR
I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA
The intensity of the waves from a point source at a distance d from the source is I. What is the intensity at a distance 2d from the source?
Answer:
The intensity at distance 2d from source is [tex]I_1 = \frac{1}{4} * I[/tex]
Explanation:
From the question we are told that
The distance of the wave from point source is d
The intensity is [tex]I[/tex]
The distance we are considering is 2d
Generally the intensity of a wave is mathematically represented as
[tex]I = \frac{ P }{\pi d^2 }[/tex]
Here P is power of point source
Now when d = 2d
[tex]I_1 = \frac{ P }{\pi (2d)^2 }[/tex]
[tex]I_1 = \frac{ 1 }{4 } * \frac{ P }{\pi d^2 }[/tex]
=> [tex]I_1 = \frac{1}{4} * I[/tex]
The intensity at a distance 2d from the source is equal to [tex]I'=\frac{I}{4}[/tex]
Given the following data:
Distance = dIntensity = ITo determine the intensity at a distance 2d from the source:
Mathematically, the intensity of a wave is given by the formula:
[tex]I=\frac{P}{\pi d^2}[/tex]
Where:
I is the intensity of a wave.P is the power.d is the distance.Since the distance is doubled (2d), we have:
Let the new intensity be [tex]I'[/tex][tex]I'=\frac{P}{\pi (2d)^2}\\\\I'=\frac{P}{4\pi (d)^2}\\\\I'=\frac{1}{4} \times \frac{P}{\pi (d)^2}\\\\I'=\frac{1}{4} \times I\\\\I'=\frac{I}{4}[/tex]
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A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb?
a. at the point, because all rays bouncing off the mirror will be parallel.
b. at the focal point of the mirror
c. at the radius of curvature of the mirror
d. none of the above, you cant make parallel rays wilth a concave mirror
Answer:
Explanation:
Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.
For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.
Therefore, the light bulb should be placed at the focal point of the mirror.
Please help!
Much appreciated!
Answer:
your question answer is 22°
The hydrogen spectrum has a red line at 656 nm, and a blue line at 434 nm. What is the first order angular separation between the two spectral lines obtained with a diffraction grating with 5000 rulings/cm?
Answer:
Explanation:
grating element or slit width a = 1 x 10⁻² / 5000
= 2 x 10⁻⁶ m
angular width of first order spectral line of wavelength λ
= λ / a
for blue line angular width
= 434 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 217 x 10⁻³ radian
for red line angular width
= 656 x 10⁻⁹ / 2 x 10⁻⁶ radian
= 328 x 10⁻³ radian
difference of their angular width
= 328 x 10⁻³ - 217 x 10⁻³
= 111 x 10⁻³ radian
Ans .
A solenoid inductor has an emf of 0.80 V when the current through it changes at the rate 10.0 A/s. A steady current of 0.20 A produces a flux of 8.0 μWb per turn.
Required:
How many turns does the inductor have?
Answer:
The number of turns of the inductor is 2000 turns.
Explanation:
Given;
emf of the inductor, E = 0.8 V
the rate of change of current with time, dI/dt = 10 A/s
steady current in the solenoid, I = 0.2 A
flux per turn, Ф = 8.0 μWb per
Determine the inductance of the solenoid, L
E = L(dI/dt)
L = E / (dI/dt)
L = 0.8 / (10)
L = 0.08 H
The inductance of the solenoid is given by;
[tex]L = \frac{\mu_o N^2 A}{l}[/tex]
Also, the magnetic field of the solenoid is given by;
[tex]B = \frac{\mu_o NI}{l}[/tex]
I is 0.2 A
[tex]B = \frac{\mu_oN(0.2)}{l} = \frac{0.2\mu_o N}{l}[/tex]
[tex]\frac{B}{0.2 } = \frac{\mu_o N}{l}[/tex]
[tex]L = \frac{\mu_o N^2 A}{l} \\\\L = \frac{\mu_o N }{l} (NA)\\\\L = \frac{B}{0.2} (NA)\\\\L = \frac{BA}{0.2} (N)[/tex]
But Ф = BA
[tex]L = \frac{\phi N}{0.2} \\\\\phi N = 0.2 L\\\\N = \frac{0.2 L}{\phi} \\\\N = \frac{0.2 *0.08}{8*10^{-6}}\\\\N = 2000 \ turns[/tex]
Therefore, the number of turns of the inductor is 2000 turns.
This question involves the concepts of magnetic flux, magnetic field, and inductance.
The inductor has "2000" turns.
The magnetic field due to an inductor coil is given as follows:
[tex]B=\frac{\mu_o NI}{L}\\\\[/tex]
where,
B = magnetic field
μ₀ = permeability of free space \
N = No. of turns
I = current = 0.2 A
L = length of inductor
Therefore,
[tex]\frac{\mu_oN}{L}=\frac{B}{0.2\ A}---------- eqn(1)[/tex]
Now, the inductance of a solenoid is given by the following formula:
[tex]E = L\frac{dI}{dt}\\\\L = \frac{E}{\frac{dI}{dt}}[/tex]
The inductance of solenoid can also be given using the following formula:
[tex]L = \frac{\mu_o N^2A}{L}[/tex]
comparing both the formulae, we get:
[tex]\frac{E}{\frac{dI}{dt}}= \frac{\mu_oN^2A}{L}\\\\E=\frac{dI}{dt}\frac{\mu_oN}{l}(NA)\\\\using\ eqn (1):\\\\E=\frac{dI}{dt}\frac{B}{0.2}(NA)\\\\[/tex]
where,
BA = magnetic flux = [tex]\phi[/tex] = 8 μWb/turn = 8 x 10⁻⁶ Wb/turn
N = No. of turns = ?
E = E.M.F = 0.8 volts
[tex]\frac{dI}{dt}[/tex] = rate of change in current = 10 A/s
Therefore,
[tex]0.8=(10)\frac{8\ x\ 10^{-6}}{0.2}N\\\\N=\frac{(0.8)(0.2)}{8\ x\ 10^{-5}}[/tex]
N = 2000 turns
Learn more about magnetic flux here:
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The attached picture shows the magnetic flux.
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastward. What are the magnitude (in T) and direction of the magnetic field at this instant?
Answer:
The values is [tex]B = 3.2 *10^{-8} \ T[/tex]
The direction is out of the plane
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 9.6 \ V/m[/tex]
The magnitude of the magnetic field is mathematically represented as
[tex]B = \frac{E}{c}[/tex]
where c is the speed of light with value
[tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]
[tex]B = 3.2 *10^{-8} \ T[/tex]
Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )