Answer:
The time taken is [tex]\Delta t = 1.5 \ s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 1.25 \ kg[/tex]
The time taken to make the first complete revolution is t= 3.60 s
The displacement of the first complete revolution is [tex]\theta = 1 rev = 2 \pi \ radian[/tex]
Generally the displacement for one complete revolution is mathematically represented as
[tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]
Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]
[tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]
[tex]\alpha = 0.9698 \ s[/tex]
Now the displacement for two complete revolution is
[tex]\theta_2 = 2 * 2\pi[/tex]
[tex]\theta_2 = 4\pi[/tex]
Generally the displacement for two complete revolution is mathematically represented as
[tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]
=> [tex]t^2 = 25.9187[/tex]
=> [tex]t= 5.1 \ s[/tex]
So
The time taken to complete the next oscillation is mathematically evaluated as
[tex]\Delta t = t_2 - t[/tex]
substituting values
[tex]\Delta t = 5.1 - 3.60[/tex]
[tex]\Delta t = 1.5 \ s[/tex]
The time for the ball to complete the next revolution is 1.5 s.
The given parameters;
mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2πThe constant angular acceleration of the ball is calculated as follows;
[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]
The time to complete the next revolution is calculated as follows;
[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]
[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]
Thus, the time for the ball to complete the next revolution is 1.5 s.
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Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
Answer:
the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle
You measure the power delivered by a battery to be 4.26 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power delivered by the battery is 17.04 W
Explanation:
Power through a circuit is given as
P = IV ....1
where P is the power
I is the current through the circuit
V is the voltage through the circuit
The voltage in a circuit is given as
V = IR ....2
Let us take the value of each resistor as equal to R
when connected in series, the total resistance will be
[tex]R_{t}[/tex] = R + R = 2R
If we assume constant voltage through the circuit, then from equation 2, the current in this case is
I = V/2R
If the resistors are connected in parallel, then the total resistance will be
[tex]\frac{1}{R_{t} }[/tex] = [tex]\frac{1}{R}[/tex] +
[tex]R_{t}[/tex] = R/2
The current in this case will be increased since the resistance is reduced
I = 2V/R
comparing the two situations, we can see that the current increased when connected in parallel to a ratio of
[tex]\frac{2V}{R}[/tex] ÷ [tex]\frac{V}{2R}[/tex] =
This means that the current increased 4 times
From equation 1, we can see that electrical power is proportional to the current at a constant voltage, therefore, the power will also increase by four times to
P = 4 x 4.26 = 17.04 W
Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s
Answer:
The time difference is 2.97 sec.
Explanation:
Given that,
Tension = 29 N
Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]
Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]
Length of first section = 2.5 m
Length of second section = 2.8 m
We need to total distance of first pulse
Using formula for distance
[tex]d=2.5+2.5[/tex]
[tex]d_{1}=5.0\ m[/tex]
We need to total distance of second pulse
Using formula for distance
[tex]d=2.8+2.8[/tex]
[tex]d_{2}=5.6\ m[/tex]
We need to calculate the speed of pulse in the first string
Using formula of speed
[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]
Put the value into the formula
[tex]v_{1}=\sqrt{\dfrac{29}{19}}[/tex]
[tex]v_{1}=1.24\ m/s[/tex]
We need to calculate the speed of pulse in the second string
Using formula of speed
[tex]v_{2}=\sqrt{\dfrac{T}}{\mu_{2}}}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{\dfrac{29}{45}}[/tex]
[tex]v_{2}=0.80\ m/s[/tex]
We need to calculate the time for first pulse
Using formula of time
[tex]t_{1}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{1}=\dfrac{5.0}{1.24}[/tex]
[tex]t_{1}=4.03\ sec[/tex]
We need to calculate the time for second pulse
Using formula of time
[tex]t_{2}=\dfrac{d_{1}}{v_{1}}[/tex]
Put the value into the formula
[tex]t_{2}=\dfrac{5.6}{0.80}[/tex]
[tex]t_{2}=7\ sec[/tex]
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=t_{2}-t_{1}[/tex]
Put the value into the formula
[tex]\Delta t=7-4.03[/tex]
[tex]\Delta t=2.97\ sec[/tex]
Hence, The time difference is 2.97 sec.
A 1000 kg car experiences a net force of 9500 N while slowing down from 30 m/s to 16 m/s. How far does it travel while slowing down?
Answer:
33.89 m
Explanation:
We must first obtain the acceleration of the car from;
F=ma
Where
F= force= 9500 N
m= mass of the car= 1000kg
a= acceleration
a= F/m= 9500/1000
a= 9.5 m/s^2
From;
V^2=u^2 + 2as
Where;
V= final velocity
u= initial velocity
s= distance covered
a= acceleration
s= v^2 -u^2/2a
s= (30)^2 -(16)^2/2×9.5
s= 900 - 256/19
s= 644/19
s= 33.89 m
The distance is 33.89 m
The first step is to calculate the acceleration
F= ma
force= 9500N
mass= 1000 kg
9500= 1000 × a
a= 9500/1000
= 9.5 m/s
v²= u² + 2as
30²= 16² + 2(9.5)(s)
900= 256 + 19s
900-256= 19s
644= 19s
s= 644/19
s= 33.89 m
Hence the distance traveled by the car is 33.89 m
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g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth and the other placed on a rocket that travels at a speed of 0.893c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to an earth-based observer
Answer:
86.4 hrs
Explanation:
The amount of bacteria is initially 1
It doubles every 24 hrs.
After first 24 hrs, the amount = 2
After next 24 hrs = 4
After next 24 hrs = 8
After next 24 hrs = 16
After next 24 hrs = 32
After next 24 hrs = 64
After next 24 hrs = 128
After next 24 hrs = 256
Total time taken to reach 256 = 24 x 8 = 192 hrs
For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship
t = t'(1 - ¥^2)^0.5
Where t is the contracted time =?
t' is the time on earth
¥ = v/c
Where v is the speed of the rocket
c is the speed of light
since v = 0.893c
¥ = 0.893
Substituting, we have
t = 192 x (1 - 0.893^2)^0.5
t = 192 x 0.2025^0.5
t = 192 x 0.45 = 86.4 hrs
an ideal gas is confined to a container with adjustable volume. the number of moles, n, and temperature, t, are constant. by what factor will the volume change if pressure increase by a factor of 5.1
Answer:
The volume will decrease by a factor of 10/51.
Explanation:
Hello,
In this case, since both moles and temperature remain constant, we can use the Boyle's law that relates the volume and pressure as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since the pressure increases by a factor of 5.1 (statement), we have:
[tex]P_2=5.1P_1[/tex]
Thus, the final volume is:
[tex]V_2=\frac{P_1V_1}{P_2} =\frac{P_1V_1}{5.1P_1}\\\\V_2=\frac{10}{51}V_1[/tex]
It means that the volume will decrease by a factor of 10/51.
Regards.
In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 540 nmnm . Part A What is the work function of this material
Answer:
Φ = 36.84 × 10^(-20) J
Explanation:
In the photoelectric effect, the energy of the incoming photon is usually used in part to extract the photoelectron from the material (work function) and then the rest is converted into kinetic energy of the photoelectron which is given by the formula;
K_max = hf - Φ
where;
hf represents the energy of the incoming photon
h is the Planck's constant
f is the light frequency
Φ is the work function of the material
K_max is the maximum kinetic energy of the photoelectrons.
From the question, we are told that no current flows unless the wavelength is less than 540 nm. This means that when the wavelength has this value, the maximum kinetic energy of the photoelectrons is zero i.e K_max = 0. Thus the energy of the incoming photons is just enough to extract the photoelectrons from the material.
Thus,
hf - Φ = 0
hf = Φ - - - (1)
We are given the wavelength as;
λ = 540 nm = 540 × 10^(-9) m
Now, let's find the frequency of the light by using the relationship between frequency and wavelength. The equation is;
f = c/λ
Where c is speed of light = 3 × 10^(8) m/s
f = (3 × 10^(8))/(540 × 10^(-9))
f = 5.56 × 10^(14) Hz
Thus, from equation 1,we can now find the work function;
Φ = hf
h is Planck's constant and has a value of 6.626 × 10^(-34) J.s
Thus;
Φ = 6.626 × 10^(-34) × 5.56 × 10^(14)
Φ = 36.84 × 10^(-20) J
White light containing wavelengths from 410 nm to 750 nm falls on a grating with 7800 slits/cm. Part APart complete How wide is the first-order spectrum on a screen 3.20 m away
Answer:
1.227 m
Explanation:
Given that
Minimum wavelength is 410 nm
Maximum wavelength is 750 nm
Grating is 7800 slits/cm
Distance is 3.2 m
To solve this question, we would use the formula
sin θ = λ/d
sin θ = (410*10^-9) / (0.01/7800)
Sin θ = 410*10^-9 / 1.282*10^-6
Sin θ = 0.32 and θ = 18.67 degrees
For the second wavelength = 750 nm
sin θ = [(0.32x750)/410]
sin θ = (240 / 410)
sin θ = 0.5853 or
θ = 35.8 degrees
And finally, the width of spectrum would be
3.2[tan 35.8 - tan 18.67]
3.2 * 0.3833
= 1.227 m
A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
An 18g bullet is shot vertically into a 10kg block. The block lifts upward 9mm. The bullet penetrates the block in a time interval of 0.001s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:
Answer:
The initial kinetic energy of the bullet is closest to 491.87 J
Explanation:
Given;
mass of bullet, m₁ = 18g = 0.018kg
mass of block, m₂ = 10kg
height moved by the block, h = 9 mm = 0.009 m
time taken for the bullet to travel through the block, t = 0.001s
let the initial velocity of the bullet = v₁
let the final velocity of the bullet = v₂
Apply the principle of conservation of linear momentum;
initial momentum = final momentum
0.018v₁ = v₂(0.018 + 10)
0.018v₁ = 10.018v₂ -----equation (1)
Apply the law of conservation of energy when the bullet lifts the block through 9mm
mgh = ¹/₂mv₂²
gh = ¹/₂v₂²
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.009)
v₂ = 0.42 m/s
Substitute in v₂ in equation 1, to determine the initial velocity of the bullet;
0.018v₁ = 10.018v₂
0.018v₁ = 10.018(0.42)
0.018v₁ = 4.208
v₁ = 4.208 / 0.018
v₁ = 233.78 m/s
Now, determine the initial kinetic energy of the bullet;
K.E₁ = ¹/₂m₁v₁²
K.E₁ = ¹/₂(0.018)(233.78)²
K.E₁ = 491.87 J
Therefore, the initial kinetic energy of the bullet is closest to 491.87 J
From a hot air balloon 2 km high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks west to see another town with angle of depression of 84 degrees. What is the distance between the two towns?
Answer:
7km
Explanation:
The angle of depression is congruent to the angle of elevation and can be explained as angle below horizontal in which the person observing an object must view for him/her to view object's that are lower than him/her.
In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.
hot air balloon 2 km high,
there exist two triangles
From trigonometry
Tanx= opposite/adjacent
Opp= 2km
Adj= X1
first triangle have base length of
Tan(16)=2/X1
X1=2/ tan(16)
X1=6.97
For Second triangle
Tanx= opposite/adjacent
Opp= 2km
Adj= X2
the other with a base length of
X2=2/tan(84)
X2=0.21
Therefore,, the total distance between the two towns is
x1+x2=6.97+0.21=7.18km
A long bar slides on two contact points and is in motion with velocity ν. A steady, uniform, magnetic field B is present. The induced current through resistor R is:
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance does the bus cover in this time frame
Answer:
Explanation:
initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s .
acceleration = (v - u) / t
= (36 - 20) / 4
a = 4 m / s ²
from the formula
v² - u² = 2 a s , s is distance covered .
putting the values
36² - 20² = 2 x 4 x s
1296 - 400 = 8 x s
s = 112 m .
Answer:112
Explanation:
PLEASE HELP ANSWER FAST As the vibration of molecules decreases, the _____ of the substance decreases. 1.temperature 2.internal energy 3.kinetic energy 4.all of the above
A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is in the xy plane. The vertical force required to lift this slab is 60 kN (F 60 kN). Find the tensions in cables DA, DB and DC (show all your workings that you do to find these)
Answer:
Fad = 28.8 kN
Fbd = 16.4 kN
Fcd = 28.1 kN
Explanation:
First, find the length of each cable.
AD = √((2 m)² + (0.5 m)² + (2.5 m)²)
AD = √10.5 m
AD ≈ 3.24 m
BD = √((1.5 m)² + (1 m)² + (2.5 m)²)
BD = √9.5 m
BD ≈ 3.08 m
CD = √((1 m)² + (1 m)² + (2.5 m)²)
CD = √8.25 m
CD ≈ 2.87 m
Next, use similar triangles to find the x, y, and z components of each tension force.
Fadx = 2/3.24 Fad = 0.617 Fad
Fady = 0.5/3.24 Fad = 0.154 Fad
Fadz = 2.5/3.24 Fad = 0.772 Fad
Fbdx = 1.5/3.08 Fbd = 0.487 Fbd
Fbdy = 1/3.08 Fbd = 0.324 Fbd
Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd
Fcdx = 1/2.87 Fcd = 0.348 Fcd
Fcdy = 1/2.87 Fcd = 0.348 Fcd
Fcdz = 2.5/2.87 Fcd = 0.870 Fcd
Now sum the forces in the x, y, and z directions:
∑Fx = ma
-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0
∑Fy = ma
-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0
∑Fz = ma
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.
-0.463 Fad + 0.811 Fbd = 0
0.811 Fbd = 0.463 Fad
Fbd = 0.571 Fad
Substitute into either of the first two equations:
-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0
-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0
-0.339 Fad + 0.348 Fcd = 0
0.348 Fcd = 0.339 Fad
Fcd = 0.975 Fad
Now substituting into the third equation:
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0
60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0
60 kN − 2.083 Fad = 0
Fad = 28.8 kN
Solving for the other two tension forces:
Fbd = 0.571 Fad = 16.4 kN
Fcd = 0.975 Fad = 28.1 kN
Answer:
Tensions of:
DA = 28.81 KN
DB = 16.45 KN
DC = 28.07 KN
Explanation:
see attached
To protect her new two-wheeler, Iroda Bike
buys a length of chain. She finds that its
linear density is 0.65 lb/ft.
If she wants to keep its weight below 1.4 lb,
what length of chain is she allowed?
Answer in units of ft.
Answer:
2.2 ft
Explanation:
0.65 lb / 1 ft = 1.4 lb / x
x ≈ 2.2 ft
A wire of 0.50m length is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.98T. The vurrent in the wire is 2.0A in the direction shown. What is the mass of the wire if the current and the magnetic field are sufficient to remove the tension in the supporting leads?
Answer:
0.1 kg or 100 g
Explanation:
The length of the wire = 0.5 m
the field magnitude = 0.98 T
the current through the wire = 2.0 A
magnetic force due to a wire carrying current is
F = [tex]IlB[/tex]
where
F is the force
[tex]I[/tex] is the current = 2 A
[tex]l[/tex] is the length of the wire
B is the magnetic field strength
Substituting, we have
F = 2 x 0.5 x 0.98 = 0.98 N
This force balances the weight of the mass
weight = mg
where m is the mass of the wire
g is acceleration due to gravity = 9.81 m/s^2
therefore, weight = m x 9.81 = 9.81m
equating this weight with the force, we have
0.98 = 9.81m
m = 0.98/9.81 = 0.099 kg ≅ 0.1 kg or 100 g
Answer:
100 g
Explanation:
A simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in a time T. The heat of fusion of this metal is
Answer:
L = Pt/M
Explanation:
Power, P= Q/t = mL/t
we know that, (Q=m×l)
Now ⇒l= Pt/M
Thus l= Pt/M
you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the winds
Answer:
Velocity
Explanation:
We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.
Hence, the given information describes wind velocity.
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
━━━━━━━━━━━━━━━━━━━━
2. The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
Answer:
We know that the maximum angle that a light ray can wake with the wall of the core is equipment to the minimum angle with the normal of the core that will give rise in total internal reflection. so using Snell's law the angle is subtracted from 90° to get the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber.
So using
n1sinစ1. = n2sinစ2
1.6sin(x1) = 1.48sin(90),
But sin(90)=1
1.6sin(စ1) = 1.48,
sin(စ1) = 1.48/1.6
စ = 68°
Explanation:
Answer:
i = 67.66⁰Explanation:
Using the Snell's law formula to solve this question which states that the ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. This constant is known as the refractive index for the given pair of media. Mathematically,
n = sin(i)/sin(r) where;
i is the angle of incidence
r is the angle of refraction.
n is the refractive index.
Given the refractive index of the optical fibre n₁ = 1.60 and that of cladding n₂ = 1.48
n₂/n₁ = sin(i)/sin(r)
The light ray can make with the wall of the core when its angle of refraction is 90⁰. The angle of incidence at this maximum point is known as the critical angle.
On substitution:
1.48/1.60 = sin(i)/sin90
1.48/1.60 = sin(i)/1
sin(i) = 1.48/1.60
sin(i) = 0.925
i = sin⁻¹0.925
i = 67.66⁰
Hence the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber is 67.66⁰.
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that oscillates in simple harmonic motion. Calculate the period of motion.
Answer:
The period of motion is 0.5 second.
Explanation:
Given;
extension of the spring, x = 3.8 cm = 0.038 m
mass of the object, m = 13 g = 0.013 kg
Determine the force constant of the spring, k;
F = kx
k = F / x
k = mg / x
k = (0.013 x 9.8) / 0.038
k = 3.353 N/m
When the object is replaced with a block of mass 20 g, the period of motion is calculated as;
[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]
Therefore, the period of motion is 0.5 second.
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth
Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
g' = 1.09 m/s²
Answer:
The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]Explanation:
Surface of earth,
[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]
free fall acceleration at height h,
[tex]g_h = \frac{GM}{(R+h)^2}[/tex]
where
G = gravitational constant
R = Radius of earth
M = mass of earth
therefore,
[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]
Where height h = 2R
[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]
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________ is a thermodynamic function that increases with the number of energetically equivalent ways to arrange components of a system to achieve a particular state.
Answer:
entropy
Explanation:
Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.
Answer and Explanation:
The computation of the shortest wavelength in the series is shown below:-
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
Where
[tex]\lambda[/tex] represents wavelength
R represents Rydberg's constant
[tex]n_f[/tex] represents Final energy states
and [tex]n_i[/tex] represents initial energy states
Now Substitute is
[tex]1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
now we will put the values into the above formula
[tex]= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )[/tex]
[tex]= 1218888.889 m^{-1}[/tex]
Now we will rewrite the answer in the term of [tex]\lambda[/tex]
[tex]\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m[/tex]
So, the whole Paschen series is in the part of the spectrum.
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
What is the radiation pressure 1.5 m away from a 700 W lightbulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.
Answer:
3.30 x 10^-7 Pascal
Explanation:
distance r = 1.5 m
power P = 700 W
the radiation pressure is given as
Pr = P/A*c
where
area of the surface A = 4πr^2
calculate for A
speed of light is c = 3×10^8 m/s
plugging above values in equation above gives
Pr = 3.30 x 10^-7 Pascal
