Answer and Explanation:
a. The computation of the percent difference between the measurements is shown below:-
The first value of g is 9.67 and the second value is 9.88
So, difference = 9.88 - 9.67
= 0.21
Percentage difference in measurement is
[tex]= \frac{0.21}{9.88}\times100[/tex]
= +/-2.13
Percent difference with 9.88
Difference = 9.88 - 9.81
= 0.07
[tex]= \frac{0.07}{9.81}\times100[/tex]
= +/-0.71%
b. The Computation of percent error of their mean is shown below:-
Mean of two values is
= [tex]\frac{9.67 + 9.88}{2}[/tex]
= 9.775
Difference = 9.81 - 9.775
= 0.035
Percentage difference is
[tex]= \frac{0.035}{9.81}\times 100[/tex]
= +/- 0.36%
At a playground, two young children are on identical swings. One child appears to be about twice as heavy as the other. Part A If you pull them back together the same distance and release them to start them swinging, what will you notice about the oscillations of the two children
Answer:
The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.
Explanation:
Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.
Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.
When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.
If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of the electric field a perpendicular distance of 0.201 m away from the center of the wire?
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
[tex]E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201} \\\\E = 0.1108 \ N/C[/tex]
Therefore, the magnitude of the electric field is 0.1108 N/C
Rice has a bulk of 0.9 g/mL, but an individual grain of rice has a density of 1.1g/mL. Water has a density of 1.0g/mL. what would happen to a cup of rice when poured into a lot of water?
Since the grains are not connected together, it's every grain for himself when they're dropped into the water.
The density of each grain is greater than the density of water, so each grain on its own sinks to the bottom.
The reason why a CUP or a BAG of rice has less density is because the grains don't all fit together perfectly, and 15% or 20% of the volume in the cup or bag is air, not rice grains.
"Rice is the perfect snack, when you're hungry for 2000 of something."
Mitch Hedberg
The mass of M1 = 12 Daltons and it has a speed of v1 = 200 m/s. The mass of M2 = 4 Daltons. What was the total momentum of the system consisting of both masses before the collision (in Dalton meters per second, assume positive to the right and negative to the left)?
Answer:
The total momentum is [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
Explanation:
The diagram illustration this system is shown on the first uploaded image (From physics animation)
From the question we are told that
The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]
The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]
The mass of the second object is [tex]M_2 = 4 \ Dalton[/tex]
The speed of the second object is assumed to be [tex]- v_2[/tex]
The total momentum of the system is the combined momentum of both object which is mathematically represented as
[tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]
substituting values
[tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]
[tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a. Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.
You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club and a golf ball.
Part A
(How fast would you need to hit the ball horizontally so it goes into a circular orbit?)
Express your answer with the appropriate units.
Part B
(If you hit the ball vertically with the same speed, to what height above you would it rise?)
Express your answer with the appropriate units.
Answer:
A) v = 1,675 10³ m / s , B) r₂ = 11,673 10⁶ m
Explanation:
A) This exercise we must use Newton's second law, where the forces of gravity are the Moon
F = m a
acceleration is centripetal
a = v² / r
force is the force of universal attraction
F = G m M / r²
we substitute
G m M / r² = m v² / r
v² = G M / r
distance
r = R_moon + h
r = 1.74 10⁶ +1.0786 10⁴
r = 1,750786 10⁶ m
we calculate
v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)
v = √ (2,8052 10⁶)
v = 1,675 10³ m / s
B) let's use energy conservation
Starting point. In the mountain
Em₀ = K + U = ½ m v² + G m M / r
Final point. Where the speed is zero
[tex]Em_{f}[/tex] = U = G mM / r₂
Em₀ = Em_{f}
½ m v² + G m M / r = G mM / r₂
1 / r₂ = (½ v₂ + G M / r) / GM
let's calculate
1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)
1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹
1 / r₂ = 8.5664 10⁻⁷
r₂ = 11,673 10⁶ m
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
a-velocity
b-mass
c-momentum
d-direction
Answer:
b. Mass
Explanation:
This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.
As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.
Therefore, the correct answer is B. Mass.
This question involves the concept of the law of conservation of momentum.
Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.
The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.
Mathematically,
[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]
where,
m₁ = mass of the first object
m₂ = mass of the second object
u₁ = velocity of the first object before the collision
u₂ = velocity of the second object before the collision
v₁ = velocity of the first object after the collision
v₂ = velocity of the second object after the collision
Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.
Learn more about the law of conservation of momentum here:
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The attached picture illustrates the law of conservation of momentum.
Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.
Answer:
T = 7.61*10^6 s
Explanation:
In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex] (1)
T: period of Mercury
r: distance between Mercury and Sun
Ms: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
You replace the values of all parameters in the equation (1):
[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]
The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high does it go
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
The ball reaches the maximum height of 54 feet
The question is about projectile motion,
the ball is shot at an angle α = 45°, and
the initial velocity u = 46 ft/s.
Under the projectile motion, the maximum height H is given by:
[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]
where, g = 9.8 m/s²
substituting the given values we get:
[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]
Hence, the maximum height is 54 feet.
Learn more:
https://brainly.com/question/2818076?referrer=searchResults
two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?
Answer:
The charge is [tex]q = 3.14 *10^{-4} \ C[/tex]
Explanation:
From the question we are told that
The mass of each ball is [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]
The distance of separation is [tex]d = 1 \ m[/tex]
Generally the weight of the each ball is mathematically represented as
[tex]W = m * g[/tex]
where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]W = 90.70 * 9.8[/tex]
[tex]W = 889 \ N[/tex]
Generally the electrostatic force between this balls is mathematically represented as
[tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]
given that the the charges are equal we have
[tex]q_1= q_2 = q[/tex]
So
[tex]F_e = \frac{k * q^2 }{d^2}[/tex]
Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force
So we have
[tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]
=> [tex]q = 3.14 *10^{-4} \ C[/tex]
The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Given data:
The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.
The distance of separation of two balls is, d = 1 m.
First of all we need to obtain the weight of ball. The weight of the ball is expressed as,
W = mg
Here,
g is the gravitational acceleration.
Solving as,
W = 90.70 × 9.8
W = 888.86 N
The expression for the electrostatic force between this balls is mathematically represented as,
[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]
Since, the charges are equal then,
[tex]q_{1} =q_{2}=q[/tex]
Also, the magnitude of force between the balls is same as the weight of one ball. Then,
F = W
Solving as,
[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Learn more about the Coulomb's law here:
https://brainly.com/question/23202809
In an RC circuit, how many time constants must elapse if an initially uncharged capacitor is to reach 80% of its final potential difference
Answer:
1.6 time constants must elapse
Explanation:
voltage on a cap, charging is given as
v = v₀[1–e^(–t/τ)]
Where R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
v = v₀[1–e^(–t/τ)]
1–e^(–t/τ) = 0.8
e^(–t/τ) = 0.2
–t/τ = –1.609
t = 1.609τ
A uniform electric field stack \rightarrow Ei subscript I with rightwards arrow on top is present in the region between infinite parallel plane plates A and B and a uniform electric field stack\rightarrow Eii subscript I I end subscript with rightwards arrow on top is present in the region between infinite parallel plane plates B and C. When the plates are vertical, stack \rightarrow Ei subscript I with rightwards arrow on top is directed to the right and stack \rightarrow Eii subscript I I end subscript with rightwards arrow on top to the left. The signs of the charges on plates A, B and C may be
a. ?, ?, ?.
b. +, ?, ?.
c. +, ?, +.
d. +, +, +.
e. any one of the above
Answer:
e. any one of the above
Explanation:
The vector indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at , its velocity and acceleration vectors are and . Which statement is correct?
Answer:
The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.
Explanation:
The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?
If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelength (as measured in water) would there be constructive interference in the light that reflects back downward
Answer:
see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
Explanation:
This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5
Let's analyze the light beam path emitted by the diver.
* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º
* also the wavelength of light in a material medium changes
λ_n = λ / n
where λ_n is the wavelength in the material and λ the wavelength in the vacuum air and n the refractive index.
If we include these aspects, the constructive interference equation is
2t = (m + ½) λ_n
2nt = (m + ½) λ
let's apply this equation to our case
λ = 2nt / (m + ½)
The incidence of replacement of the oil with respect to water is
n = n_oil / n_water = 1.5 / 1.33
n = 1,128
let's calculate
λ = 2 1,128 t / (m + ½)
λ = 2,256 t / (m + ½)
In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation
t = 0.001 mm = 1 10⁻⁶ m
the formula remains
λ = 2,256 10⁻⁶ / (m + ½)
Let's find what values of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m
m + ½ = 2,256 10⁻⁶ / λ
m = 2,256 10⁻⁶ / λ - ½
light purple lan = 400 10⁻⁹m
m = 2,256 10-6 / 400 10⁻⁹ - ½
m = 5.64 - 0.5
m = 5.14
m = 5
red light λ = 700 10⁻⁹m
m = 2,256 1-6 / 700 10⁻⁹ - ½
m = 3.22 - 0.5
m = 2.72
m = 3
we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?
Answer:
29.96cm
Explanation:
Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.
Now using the lens equation as follows;
[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex] -------------(i)
Where;
f = focal length of the lens
v = image distance as seen by the lens
u = object distance from the lens
From the question;
v = -151cm [-ve since the image formed is virtual]
u = 25cm
Rewrite equation (i) to have;
[tex]f = \frac{uv}{u+v}[/tex]
Substitute the values of v and u into the equation;
[tex]f = \frac{25*(-151)}{25-151}[/tex]
[tex]f = \frac{-3775}{-126}[/tex]
f = 29.96cm
The focal length should be 29.96cm
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 28.0° above the horizontal. The car accelerates uniformly to a speed of 2.35 m/s in 14.0 s and then continues at constant speed.(A) What power must the winch motor provide when the car is moving at constant speed? kW(B) What maximum power must the motor provide? kW(C) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m?
Answer:
a) P = 10.27 kW
b) Pmax = 10.65 kW
c) E = 5.47 MJ
Explanation:
Mass of the loaded car, m = 950 kg
Angle of inclination of the shaft, θ = 28°
Acceleration due to gravity, g = 9.8 m/s²
The speed of the car, v = 2.35 m/s
Change in time, t = 14.0 s
a) The power that must be provided by the winch motor when the car is moving at constant speed.
P = Fv
The force exerted by the motor, F = mg sinθ
P = mgv sinθ
P = 950 * 9.8 *2.35* sin28°
P = 10,271.3 W
P = 10.27 kW
b) Maximum power that the motor must provide:
[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]
c) Total energy transferred:
Length of the track, d = 1250 m
[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]
A girl walks East at 2.7 m/s. What is the y component of her velocity?
Answer:
The y component of her velocity is zero
Explanation:
Given that North and South represent the positive and negative y component of the velocity vector.
And East and West represent the positive and negative x component of the velocity vector.
For a girl walking east at 2.7 m/s
The x component of velocity is;
vx = 2.7 m/s (east is positive x axis)
The y component of velocity is;
vy = 0 m/s (no movement in the northern and southern direction)
Therefore, The y component of her velocity is zero (0m/s)
A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the power exerted by the student
Answer:
The power exerted by the student is 51.2 W
Explanation:
Given;
extension of the elastic band, x = 0.8 m
time taken to stretch this distance, t = 0.5 seconds
the spring constant, k = 40 N/m
Apply Hook's law;
F = kx
where;
F is the force applied to the elastic band
k is the spring constant
x is the extension of the elastic band
F = 40 x 0.8
F = 32 N
The power exerted by the student is calculated as;
P = Fv
where;
F is the applied force
v is velocity = d/t
P = F x (d/t)
P = 32 x (0.8 /0.5)
P = 32 x 1.6
P = 51.2 W
Therefore, the power exerted by the student is 51.2 W
A 100 cm length of nichrome wire has a radius of 0.50 mm, a resistivity LaTeX: \rho_0ρ 0= 1.0 × 10-6 Ω ∙ m , and a temperature coefficient LaTeX: \alphaα = 0.4 × 10-3 (oC)-1. At T0 = 20 oC the wire carries current of 0.50 A. How much power does the wire dissipate at a temperature T = 350 oC? Assume the potential difference across the ends of the wire remains constant. Group of answer choices
Answer:
P₃₅₀ = 0.28 watt
Explanation:
First we find the resistance of the wire at 20°C:
R₀ = ρL/A
where,
ρ = resistivity = 1 x 10⁻⁶ Ωm
L = Length of wire = 100 cm = 1 m
A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²
Therefore,
R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)
R₀ = 1.27 Ω
Now, from Ohm's Law:
V = I₀R₀
where,
V = Potential Difference = ?
I₀ = Current Passing at 20°C = 0.5 A
Therefore,
V = (0.5 A)(1.27 Ω)
V = 0.64 volts
Now, we need to find the resistance at 350°C:
R₃₅₀ = R₀(1 + αΔT)
where,
R₃₅₀ = Resistance at 350°C = ?
α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹
ΔT = Difference in Temperature = 350°C - 20°C = 330°C
Therefore,
R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]
R₃₅₀ = 1.44 Ω
Now, for power at 350°C:
P₃₅₀ = VI₃₅₀
where,
P₃₅₀ = Power dissipation at 350°C = ?
V = constant potential difference = 0.64 volts
I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)
Therefore,
P₃₅₀ = V²/R₃₅₉
P₃₅₀ = (0.64 volts)²/(1.44 Ω)
P₃₅₀ = 0.28 watt
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What are the two types of long-term memory?
A. sensory and short-term
B. iconic and echoic
C. explicit and implicit
D. recency and primacy
Answer:
C. explicit and implicit
Explanation:
E20
Determine the magnitude of the force between two 11 m-long parallel wires separated by 0.033 m, both carrying 5.2 A in the same direction.
Answer:
[tex]F=1.8\times 10^{-3}\ N[/tex]
Explanation:
We have,
Length of wires is 11 m
Separation between wires is 0.033 m
Current in both the wires is 5.2 A
It is required to find the magnitude of force between two wires. The force between wires is given by :
[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]
So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]
If you were to experimentally determine the length of the pendulum, why would you not get the same length in Iowa?
Answer:
The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.
Explanation:
The period of oscillation is calculated as;
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]
where;
L is the length of the pendulum bob
g is acceleration due to gravity
If we make L the subject of the formula in the equation above, we will have;
[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]
The length of the pendulum depends on acceleration due to gravity (g).
Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.
in which direction does a buoyant force act on an object?
A. Right or Left
B. Down
C. Opposite to it's motion
D. Up
A buoyant force always acts up on an object. (D)
You are trying to overhear a juicy conversation, but from your distance of 20.0 m, it sounds like only an average whisper of 20.0 dB. So you decide to move closer to give the conversation a sound level of 70.0 dB instead. How close should you come?
Given that,
Distance = 20.0 m
Average whisper = 20.0 dB
Sound level = 70.0 dB
We know that,
The minimum intensity is
[tex]I_{o}=10^{-12}\ W/m^2[/tex]
We need to calculate the sound intensity in the distance of 20 m
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{a}}{I_{o}})[/tex]
Put the value into the formula
[tex]20=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{2}=\dfrac{I_{a}}{10^{-12}}[/tex]
[tex]I_{a}=10^{-10}\ W/m^2[/tex]
If the conversation a sound level of 70.0 dB instead
We need to calculate the sound intensity
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{b}}{I_{o}})[/tex]
Put the value into the formula
[tex]70=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{7}=\dfrac{I_{b}}{10^{-12}}[/tex]
[tex]I_{b}=10^{-5}\ W/m^2[/tex]
We know that,
The intensity is inversely proportional with the square of the distance.
We need to calculate the distance
Using formula of intensity
[tex]\dfrac{I_{a}}{I_{b}}=\dfrac{R_{b}^2}{R_{a}^2}[/tex]
Put the value into the formula
[tex]\dfrac{10^{-10}}{10^{-5}}=\dfrac{R_{b}^2}{20^2}[/tex]
[tex]R_{b}^2=20^2\times\dfrac{10^{-10}}{10^{-5}}[/tex]
[tex]R_{b}=\sqrt{20^2\times\dfrac{10^{-10}}{10^{-5}}}[/tex]
[tex]R_{b}=0.063\ m[/tex]
Hence, The distance from the conversation should be 0.063 m.
The distance you should move to achieve a loudness of 70 dB is; 0.0633 m
We are given;
Average distance 1; R_1 = 20 m
Average whisper 1; dB1 = 20 dB
Average whisper 2; dB2 = 70 dB
Now, the formula for loudness in dB is;
dB = 10log(I/I_o)
Where;
I is the intensity of the sound
I_o is the minimum intensity that a human ear can detect = 10^(-12) W/m²
Thus, for dB1 = 20 dB;20 = 10log (I/10^(-12))
20/10 = log (I/10^(-12))
log (I/10^(-12)) = 2
10² = (I/10^(-12))
I = 10² × 10^(-12)
I_1 = 10^(-10) W/m²
Similarly, for dB2 = 70 dB;70 = 10log (I/10^(-12))
70/10 = log (I/10^(-12))
log (I/10^(-12)) = 7
10^(7) = (I/10^(-12))
I = 10^(7) × 10^(-12)
I_2 = 10^(-5) W/m²
The relationship between their intensities and distance is;
I_1/I_2 = (R_2/R_1)²
Where R_2 is the distance you should move to achieve a loudness of 70 dB.
Thus;
(10^(-10))/(10^(-5)) = R_2/20
(R_2)/20 = √(10^(-5))
(R_2)/20 = 0.00316227766
R_2 = 20 × 0.00316227766
R_2 = 0.0632 m
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Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diameter of the 25th ring.
Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
Diameter of [tex]25^{th}[/tex] Newton Ring = 0.97 cm
Newton Rings is an experiment based on principle of thin film interference
In Newton Rings Experiment the Diameter of [tex]n^{th}[/tex] dark ring is given by equation (1)
[tex]\rm D_n= 2\sqrt{n\lambda R} ......(1)\\where \; \\D_n = Diameter\; of \; n^{th} \; dark \; ring }\\n = Number \; of \; Ring\\\lambda = Wavelength \\R = Radius \; of \; Curvature \; of\; the \; lens[/tex]
From the condition given
[tex]\rm D_5 = 0.3 \; cm \\D_{15} = 0.62 \; cm\\\\D_{25} = To \; be \; determined \\[/tex]
Putting the values in equation (1) for fifth diameter we get
[tex]\rm D_5 = 0.3=2\sqrt{5\lambda R}.......(2) \\D_{15} = 0.62 = 2\sqrt{15\lambda R}.......(3) \\\\Equation \; (3) - Equation (2) \\\\0.32 = 2\sqrt{\lambda R} ( \sqrt{15} -\sqrt{5})\\\\2\sqrt{\lambda R } = 0.1954....(4)\\[/tex]
So From equation (1) and (4)
[tex]\rm Diameter \; of \; 25^{th} Ring =D_{25} = 2\sqrt{\lambda R } \times \sqrt{25} \\\\D_{25} = 0.1954\times 5 \\\\D_{25} = 0.97 \; cm[/tex]
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Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless. they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.
Required:
a. What is the energy, in joules, of the photon?
b. What is the energy, in electron volts. of the photon?
Answer:
a. E = 1.02*10^-27 J
b. E = 6.39*10^-9eV
Explanation:
a. In order to calculate the energy of the radio photon, you use the following formula:
[tex]E=hf[/tex] (1)
h: Planck's constant = 6.626*10^-34 Js
f: frequency of the photon = 1545kHz = 1.545*10^6 Hz
Then, by replacing you obtain the energy of the photon:
[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]
b. In electron volts, the energy of the photon is:
[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]
I have a question regarding friction in rolling without slipping.
There is a disk that is being acted by a force at an angle. I am asked to find the acceleration of center of mass and angular acceleration. The problem specifies, “there is friction between the disk and the ground, but the disk rolls without slipping”.
What does that mean? Does it mean I can ignore friction and use F as the only force producing torque? Does friction produce torque in this case?
I’ll leave a picture here:
Explanation:
They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.
(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)
In this case, you can't ignore friction because the disk wouldn't roll without it.
As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.
(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)
Hopefully this helps clear up the misunderstanding for you.
How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller
Answer:
The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia = [tex]I[/tex]
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = [tex]5I[/tex] (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
therefore,
for the first case, the K.E. is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
and for the second case, the K.E. is given as
[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]
[tex]KE = \frac{1}{10}Iw^{2}[/tex]
this is one-tenth the kinetic energy before its spinning characteristics were changed.
This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.
Relation between Kinetic energy and Moment of Inertia:
Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.Now, let's consider moment of inertia = I and angular speed = ω
It is asked that what would be change in Kinetic energy if
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
The kinetic energy of a spinning body is given as:
[tex]K.E.=\frac{1}{2} I. w^2[/tex]
On substituting the values, we will get:
[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]
Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.
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