Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²
[tex]A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m[/tex]
(c) the maximum speed attained by the object during its motion
[tex]E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s[/tex]
a. The total energy of the toy at any point in its motion is 0.0416 Joules.
b. The amplitude of the motion is equal to 0.0167 meter.
c. The maximum speed attained by the toy during its motion is 0.577 m/s.
Given the following data:
Mass of toy = 0.250 kgSpring constant = 300 N/mDistance = 0.0120 mSpeed = 0.400 m/sa. To find the total energy of the toy at any point in its motion:
Mathematically, the total energy of an object undergoing simple harmonic motion (SHM) is given by:
[tex]E = \frac{1}{2} MV^2 + \frac{1}{2} kx^2[/tex]
Where:
k is the spring constant.x is the distance.M is the mass of an object.V is the speed of an object.Substituting the given parameters into the formula, we have;
[tex]E = \frac{1}{2} \times 0.25 \times 0.400^2 + \frac{1}{2} \times 300 \times 0.0120^2\\\\E = 0.125 \times 0.16 + 150 \times 0.000144\\\\E=0.02+0.0216[/tex]
E = 0.0416 Joules.
b. To find the amplitude of the motion, we would use this formula:
[tex]A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2 \times 0.0416}{300} } \\\\A = \sqrt{2.77 \times 10^{-4}}[/tex]
A = 0.0167 meter.
c. To find the maximum speed attained by the object during its motion:
[tex]V_{max} = \sqrt{\frac{2E}{M} } \\\\V_{max} = \sqrt{\frac{2 \times 0.0416}{0.25} } \\\\V_{max} = \sqrt{2.77 \times 10^{-4}}[/tex]
Maximum speed = 0.577 m/s
Read more: https://brainly.com/question/14621920
What is the Malebioncy of a Capacitor?
Answer:
The switching rate between the steady state and the normal state of a capacitor
Explanation:
This was a hard one! Could only find it in my textbook. Anyways this basically is the rate which the capacitors switches back from steady state and normal state from when it charges and discharges over time. This has many purposes as a special type of diode or any other transistor type device etc etc.
You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over
the 12.0 km drive will be
Explanation:
average speed = total distance travelled / total time travelled
time to travel the first 6km: 6 / 50 = 3/25 (h)
time to travel the next 6km: 6 / 90 = 1/15 (h)
[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]
The total time: 3/25 + 1/15 = 14/75 (h)
Average speed over the 12 km drive will be:
[tex] \frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1} [/tex]
4. How does the type of medium affect a sound wave?
Answer:
The type of medium affects a sound wave as sound travels with the help of the vibration in particles.
Explanation:
As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.
The speed of sound in a given medium is determined by its density and stiffness (or compressibility in the case of gases).The speed of sound increases with the rigidity (or lack of compressibility) of the medium. The speed of sound decreases with increasing medium density.
What type of medium affect a sound wave?Any material or area through which a wave is transmitted is referred to as a medium. Four variables impact a wave's speed: wavelength, frequency, medium, and temperature. The wavelength and frequency are multiplied to determine the wave speed (speed = l × f).
Therefore, The rate at which energy is transferred through a medium depends on the amplitude of the vibrations of its constituent particles; the higher this rate, the more powerful the sound wave.
Learn more about sound wave here:
https://brainly.com/question/21995826
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A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for:
a. its speed
b. its mass.
Answer:
Given that K.E is
1/2mv²
So to find speed v,
Make it subject
K.E= 1.2mv²
However radial force = magnetic force
So mv²/r= qvB
So v subject
V= 2K.E/ qBr that is speed
To find mass
K.E = 1/2mv²
Puy value of v
So KE= 1/2m(2K.E/qBr)
m= (qBr)/2K.E
That is mass
Answer:
m = qbr/v
v = 2k/qbr
Explanation:
When a charged particle enters a magnetic field, it experiences a force that is always perpendicular to the velocity. This force provides a centripetal force, and thus, we have
qvb = mv²/r
if we make m the subject of the formula, we will have
m = qbr/v
Recall that the kinetic energy, KE = ½mv²
Now, let's make v² the subject of formula, we have
v² = 2K/m
now, we substitute for m from the equation we got earlier
v² = 2K / (qbr/v)
v² = 2Kv / qbr, if we simplify further, we have
v = 2k / qbr
Therefore, we can say that the expression for the mass and speed is respectively,
m = qbr/v
v = 2k/qbr
6. Solve (5.87 x 10^7)(4.200 x 10^11). Be
sure your answer is in scientific notation.
Round to two decimal places.
Explanation:
We need to solve [tex](5.87\times 10^7)(4.2\times 10^{11})[/tex]
Firstly, multiplying 5.87 and 4.2 = 24.654
Now taking exponent of 10.
We know that : [tex]x^a{\cdot} x^b=x^{a+b}[/tex]
It means, [tex]10^7{\cdot} 10^{11}=10^{11+7}=10^{18}[/tex]
So,
[tex](5.87\times 10^7)(4.2\times 10^{11})=24.654\times 10^{18}[/tex]
In scientific notation,
[tex](5.87\times 10^7)(4.2\times 10^{11})=2.4654\times 10^{19}[/tex]
Hence, the value of [tex](5.87\times 10^7)(4.2\times 10^{11})[/tex] is [tex]2.4654\times 10^{19}[/tex]
Answer:
Explanation:
We need to solve
Firstly, multiplying 5.87 and 4.2 = 24.654
Now taking exponent of 10.
We know that :
It means,
So,
In scientific notation,
Hence, the value of is
Select the correct answer. Physics is explicitly involved in studying which of these activities? A. the mixing of metals to form an alloy B. the metabolic functions of a living organism C. the motion of a spacecraft under gravitational influence D. the depletion of the atmospheric ozone layer due to pollutants E. the killing of cancerous cells by radiation therapy
Answer:
C. the motion of a spacecraft under gravitational influence
Which is the property of mattter in which substance can transfer heat to electricity
Conductivity is the property of matter in which a substance can transfer heat or electricity.
Additional information:-Matter : Anything which occupies space and has mass is called matter.
Chemical classifications
Pure Substances ( made of one kind of substance )Impure Substances ( mixture )Physical classifications
Solid Liquid GasPlasma ( made of ions and free electrons )BEC ( Bose Einstein Condensate )Fermionic Condensate ( It discovered in 2003 )What is the shortest possible time in which a bacterium to travel distance of 8.4cm across a Petri dish at a constant velocity of 1.2 cm/s
Answer:
[tex] \boxed{\sf Shortest \ possible \ time = 7 \ seconds} [/tex]
Given:
Distance travelled (s) = 8.4 cm
velocity (v) = 1.2 cm/s
To Find:
Shortest possible time (t) in which a bacterium travel a distance 8.4 cm across a Petri Dish
Explanation:
[tex] \boxed{ \bold{\sf Time \ (t) = \frac{Distance \ travelled \ (s)}{Velocity \ (v)}}}[/tex]
Substituting values of Distance travelled (s) & Velocity (v) in the equation:
[tex] \sf \implies t = \frac{8.4}{1.2} [/tex]
[tex] \sf \implies t = \frac{7 \times \cancel{1.2}}{ \cancel{1.2}} [/tex]
[tex] \sf \implies t = 7 \: s[/tex]
A parachuter, falling with a constant speed, drops 456m in 14.4s.
Determine all unknowns.
S =
51
m/s
d =
456
m
t =
14.4
S
Answer:
The parachuter is falling at a speed of 31.667 meters per second.
Explanation:
Given that parachuter falls at constant speed and travelled distance and time are known, the unknown is speed, measured in meters per second, which is obtained by the following kinematic expression:
[tex]s = \frac{d}{t}[/tex]
Where:
[tex]d[/tex] - Travelled distance, measured in meters.
[tex]t[/tex] - Time, measured in seconds.
If [tex]d = 456\,m[/tex] and [tex]t = 14.4\,s[/tex], the speed of the parachuter is:
[tex]s = \frac{456\,m}{14.4\,s}[/tex]
[tex]s = 31.667\,\frac{m}{s}[/tex]
The parachuter is falling at a speed of 31.667 meters per second.
Suppose no stars more massive than about 2 solar masses had ever formed. Would life as we know it have been able to develop
Answer:
No, life would not be to develop
Explanation:
Stars less massive than about 2 solar masses can only produce natural element up to carbon and oxygen, which are the basic elements for building life. However, other more massive elements are needed by life to thrive and function properly, more massive elements like phosphorus, iron necessary for oxygen circulation, calcium for a strong support system, and silicon are essential for for life to form and be sustained here on Earth.
What is force? What creates it?
Answer:
its an interaction that can move an object; push or pull makes it or gravity, magnetism
Explanation:
its all in the answer
Answer:
In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object withmass to change its velocity (which includes to begin moving from a state of rest), i.e., toaccelerate.
Two vehicles collide and stick together. After the collision, their combined y-momentum is 2.40 × 104 kilogram meters/second, and their x-momentum is 7.00 × 104 kilogram meters/second. What is the angle of the motion of the two vehicles, with respect to the x-axis?
Explanation:
It is given that,
Momentum in y direction is [tex]2.4\times 10^4\ kg-m/s[/tex]
Momentum in x direction is [tex]7\times 10^4\ kg-m/s[/tex]
We need to find the angle of the motion of the two vehicles, with respect to the x-axis. The angle between two vectors is given by :
[tex]\tan\theta=\dfrac{p_y}{p_x}\\\\\tan\theta=\dfrac{2.4\times 10^4}{7\times 10^{4}}\\\\\theta=\tan^{-1}\left(0.342\right)\\\\\theta=18.88^{\circ}[/tex]
So, the angle of the motion of the two vehicles is 18.88 degrees.
Find the sum of the following vectors A=3i-12j and B=4i+7j
Answer:
(I). The sum of the vectors is (7i-5j).
(II). The sum of the vectors is (8i+7j).
Explanation:
Given that,
(I). Vector A [tex]A=3i-12j[/tex]
Vector B [tex]B=4i+7j[/tex]
Suppose, (II). Vector A [tex]A=6i+15j[/tex]
Vector B [tex]B=2i-8j[/tex]
(I). We need to calculate the sum of the vectors
Using formula of sum
[tex]\vec{C}=\vec{A}+\vec{B}[/tex]
Where,
[tex]\vec{A}= vector A[/tex]
[tex]\vec{B}= vector B[/tex]
[tex]\vec{C}= sum of the vector A and b
Put the value into the formula
[tex]\vec{C}=(3i-12j)+(4i+7j)[/tex]
[tex]\vec{C}=7i-5j[/tex]
(II). We need to calculate the sum of the vectors
Using formula of sum
[tex]\vec{C}=\vec{A}+\vec{B}[/tex]
Put the value into the formula
[tex]\vec{C}=(6i+15j)+(2i-8j)[/tex]
[tex]\vec{C}=8i+7j[/tex]
Hence, The sum of the vectors is (7i-5j).
The sum of the vectors is (8i+7j).
If you unbend a paper clip made from 1.5 millimeter diameter wire and push one end against the wall, what force must you apply to give a pressure of 120 atmospheres
Answer:
The force is [tex]F = 21.48 \ N[/tex]
Explanation:
From the question we are told that
The diameter of the wire is [tex]d = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
The pressure is [tex]P = 120 \ a.t.m = 120 * 101.3 *10^{3} = 12156000 Pa[/tex]
Generally the radius of the of the wire is
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{ 1.5 *10^{-3}}{2}[/tex]
=> [tex]r = 7.5 *10^{-4} \ m[/tex]
The Area is evaluated as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * 7.5 *10^{-4}[/tex]
=> [tex]A = 1.7673*10^{-6} \ m^2[/tex]
Generally pressure is mathematically represented as
[tex]P = \frac{F}{A }[/tex]
=> [tex]F = P* A[/tex]
=> [tex]F = 12156000 * 1.767*10^{-6}[/tex]
=> [tex]F = 21.48 \ N[/tex]
A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?
Answer:
0.182C
Explanation:
Using Q= It
= 13x10^3 . 14x10^-6
= 0.182C
If you are driving 95 km????h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period?
Answer:
52.7 m
Explanation:
Given that
speed of the vehicle, v = 95 km/h
time of inattentiveness, t = 2 s
distance travelled, s = ?
Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.
95 km/h = 95 * 1000/3600 m/s
95 km/h = 95000/3600 m/s
95 km/h = 26.38 m/s
Now, we use our derived speed in m/s
Speed of a moving vehicle is given by,
v = s/t, where
v = speed in m/s
s = distance travelled, in m
t = time spent, in s
if we make d the subject of formula by rearranging the equation, we have
s = v * t
distance travelled, s = 26.38 * 2
distance travelled, s = 52.7 m
therefore, during this inattentive period, 52.7 m was travelled.
A large number of very industrious people make a very long pole. It is 10.0 light years long! ( As they measure it. ) Soon a spaceship flies along the length of the pole at 90% the speed of light. How much time passes on the spaceship from the moment the ship passes the first end of the pole to the moment the ship passes the second end of the pole
Answer:
L = L0 ( 1 - v^2/c^2))1/2 where L0 is the proper length
L = 10 L-y (1 - .9^2)^1/2 = 4.36 L-y length of pole measured by ship
t = 4.36 L-y / .9 c = 4.84 y since the ship travels at .9 c
Each wheel of a 320 kg motorcycle is 52 cm in diameter and has rotational inertia 2.1 kg m2 . The cycle and its 75 kg rider are coasting at 85 km/h on a flat road when they encounter a hill. If the cycle rolls up the hill with no applied power and no significant internal friction, what vertical height will it reach
Answer:
The value is [tex]h = 32.91 \ m[/tex]
Explanation:
From the question we are told that
The diameter of each wheel is [tex]d = 52 \ cm = 0.52 \ m[/tex]
The mass of the motorcycle is [tex]m = 320 \ kg[/tex]
The rotational kinetic inertia is [tex]I = 2.1 \ kg \ m^2[/tex]
The mass of the rider is [tex]m_r = 75 \ kg[/tex]
The velocity is [tex]v = 85 \ km/hr = 23.61 \ m/s[/tex]
Generally the radius of the wheel is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.52}{2}[/tex]
=> [tex]r = 0.26 \ m[/tex]
Generally from the law of energy conservation
Potential energy attained by system(motorcycle and rider ) = Kinetic energy of the system + rotational kinetic energy of both wheels of the motorcycle
=> [tex]Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} Iw^2 + \frac{1}{2} Iw^2[/tex]
=> [tex]Mgh = \frac{1}{2} * Mv^2 + Iw^2[/tex]
Here [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v }{r }[/tex]
So
[tex]Mgh = \frac{1}{2} * Mv^2 + I \frac{v}{r} ^2[/tex]
Here [tex]M = m_r + m[/tex]
[tex]M = 320 + 75[/tex]
[tex]M = 395 \ kg[/tex]
[tex]395 * 9.8 * h = 0.5 * 395 * (23.61)^2 + 2.1 *[\frac{ 23.61}{ 0.26} ] ^2[/tex]
=> [tex]h = 32.91 \ m[/tex]
Which statement about the ocean is true? A. No evaporation or precipitation in the water cycle occurs over the ocean. B. Most evaporation and precipitation in the water cycle occur over the ocean. C. All evaporation and precipitation in the water cycle occur over the ocean. D. Evaporation, but not precipitation, in the water cycle occurs over the ocean.
Answer:
A
Explanation:
Answer:
A
Explanation:
No evaporation or precipitation in the water cycle occurs over the ocean.
The pressure at the bottom of a full barrel of water is Poriginal . Determine what happens to the pressure when the radius or height of the barrel is changed and water is added to make the barrel full again.
Answer:
a) P' = P_original, b) P ’= P_original + ρ g Δh
Explanation:
The expression for nanometric pressure is
P = ρ g h
where ρ is the density of the liquid and h is the height
a) we change the radius of the barrel, but keeping the same height
as the pressure does not depend on the radius it remains the same
P' = P_original
b) We change the barrel height
h ’≠ h
we substitute in the equation
P ’= ρ g h’
h ’= h + Δh
P ’= ρ g (h + Δh)
P ’= (ρ g h) + ρ g Δh
P ’= P_original + ΔP
In this case, the pressure changes due to the new height,
*if it is higher than the initial one, the pressure increases
*if the height is less than the initial one, the pressure is less
A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determine the corresponding magnitude of force P.
Answer:
[tex]F_x=208.25\ N[/tex]
Explanation:
Given that,
Mass of a crate is 22 kg
It moved up along the 15 degrees incline without tipping.
We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.
It means that the horizontal component of force is given by :
[tex]F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N[/tex]
So, the horizontal component of force is 208.25 N.
What is the heat-loss rate through the slab if the ground temperature is 5 ∘C while the interior of the house is 25 ∘C?
Complete question :
A 12 m x 15 m house is built on a 12-cm-thick concrete slab.
What is the heat-loss rate through the slab if the ground temperature is 5°C while the interior of the house is 25°C
Answer:
3kW
Explanation:
Given the following :
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
t in metres :
100cm = 1m
12cm = (12/100)m
= 0.12m
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
Thermal conductivity of concrete (K) is approximately 1 Wm/k
Using the relation:
Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W = 3kW
The heat-loss rate is 3kW
Given that,
Dimension of house :
Length = 12m
Width = 15m
Thickness of concrete slab (t) = 12cm
We know that
100cm = 1m
so,
12cm = (12/100)m
= 0.12m
And,
Ground temperature (Tg) = 5°C
Interior temperature = (Th) = 25°C
calculation of heat loss rate:Q = KA * [ (Th - Tg) / d]
A = Length * width = (12 *15) = 180
Q = (1 * 180) * [(25°C - 5°C) / 0.12]
Q = 180 * (20/0.12)
Q = 180 * 16.6666
Q = 3,000W
= 3kW
learn more about the temperature here: https://brainly.com/question/16940730
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
pls answer quickly. Thanks
Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.420 with the floor. If the train is initially moving at a speed of 57.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor
Answer:
The distance is [tex]s= 30.3 \ m[/tex]
Explanation:
From the question we are told that
The coefficient of static friction is [tex]\mu_s = 0.42[/tex]
The initial speed of the train is [tex]u = 57 \ km /hr = 15.8 \ m/s[/tex]
For the crate not to slide the friction force must be equal to the force acting on the train i.e
[tex]-F_f = F[/tex]
The negative sign shows that the two forces are acting in opposite direction
=> [tex]mg * \mu_s = ma[/tex]
=> [tex]-g * \mu_s = a[/tex]
=> [tex]a = -9.8 * 0.420[/tex]
=> [tex]a = -4.116 m/s^2[/tex]
From equation of motion
[tex]v^2 = u^2 + 2as[/tex]
Here v = 0 m/s since it came to a stop
=> [tex]s= \frac{v^2 - u^2 }{ 2 a}[/tex]
=> [tex]s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}[/tex]
=> [tex]s= 30.3 \ m[/tex]
Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction, n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minumim film thickness will result in minimum reflection of this light?A. lambda/(4*n2)B. lambda/n2C. lambda/4D. lambda(2*n1)E. lambdaF. lambda/(2*n2)G. lambda/n1H. lambda/(4n1)I. lambda/2
Answer:
The correct option is H
Explanation:
From the question we are told that
The index of refraction of coating is [tex]n_1[/tex]
The index of refraction of material is [tex]n_2[/tex]
Generally the condition for constructive for a thin film interference is mathematically represented
[tex]2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }[/tex]
Here t represents the thickness
For minimum thickness m = 0
So
[tex]2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }[/tex]
=> [tex]t =\frac{\lambda}{4n_1 }[/tex]
Assume that helium behaves as an ideal monatomic gas. If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
6235.5J
Explanation:
Using ( nစ)p= ncp x change in temp
But cp= ( 1+ f/2)R
So cp= ( 1+ 3/2R
Cp= 5R/2
So = n x 5R/2x 150k
= 2 x 5/2x 8.314 x150
= 6235.5J
Answer:
2500 J
Explanation:
Q=(3/2)nRΔT
Q=(3/2)*2 mol*(8.314 J/mol*k)*100 k
Q=2494 J
uncertainty propagation question #2
Hi all, I am trying to calculate the uncertainty and volume for a rectangular block with the measurements being 8.7cm, 5.2cm, 5.4cm. I am struggling with the uncertainty propagation, and I am unsure if I did this correctly. Heres what I've tried.
I found the uncertainty for each individual measurement to be .1 because they all have 1 decimal place. I Added this to the formula with the measurements, took the square root of the sum of the squares with the uncertainty for each individual measurement being in the numerator and the measurement in the denominator, as follows: √(.1/8.7)\^2 + (.1/5.2)\^2 + (.1/5.4)\^2. My final answer was volume= 244.296 +/- .029 cm\^3. I rounded the uncertainty that I got from the equation to 2 significant figures because that’s what the smallest measurement has. Did I do this correctly?
Answer:
The correct treatment of uncertainties for the volume is shown below
Explanation:
In order to estimate the uncertainty in the volume which is derived via the formula:
[tex]V = w*l*h[/tex]
you normally start with the relative errors [tex](\frac{\delta Q}{Q})[/tex] of each quantity (Q) measured, since they are so easy to handle, stating that the relative error in the Volume is the addition of the relative errors in each quantity:
[tex]\frac{\delta V}{V} =\frac{\delta w}{w} +\frac{\delta l}{l} +\frac{\delta h}{h}[/tex]
and finally solve for [tex]\delta V[/tex] by multiplying both sides by the volume you calculated.
In your case, this becomes:
[tex]\delta V =V \left \{\frac{\delta w}{w} +\frac{\delta l}{l} +\frac{\delta h}{h}\right \} \\\delta V = 244.296 \left \{\frac{0.1}{5.4} +\frac{0.1}{8.7} +\frac{0.1}{5.2}\right \}\\\delta V = 244.296 \, (0.04924354)\\\delta V = 12.03 \,\,cm^3[/tex]
Then, since the standard practice is to write the uncertainty with ONLY ONE significant figure, the rounding of your uncertainty becomes:
[tex]\delta V=10\,\,cm^3[/tex]
Giving this, you need to express the final measurement as:
[tex]V=240\,\,cm^3\,+/- 10 \,\,cm^3[/tex]
making sure that the expression for the volume doesn't have significant figures passed the limitation imposed by its uncertainty (in this case the tenths).
Please notice as well that in the treatment you did, you:
1) ended up with an uncertainty even smaller than the relative uncertainty of each measurement (which cannot be possible since relative uncertainties add-up)
2) are not rounding your uncertainty to ONE SIG FIG.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 5.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m=0 and m=1 maxima to be 35 cm. What slit separation is required in order to produce the desired interference pattern?
Answer:
The distance of separation is [tex]d = 9.04 *10^{-6 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 633\ nm = 633 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 5.0 \ m[/tex]
The distance between the fringes is [tex]y = 35 \ cm = 0.35 \ m[/tex]
Generally the distance between the fringes is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
Here d is the distance of separation between the slit
=> [tex]d = \frac{ \lambda * D }{y }[/tex]
=> [tex]d = \frac{ 633 *10^{-9} * 5 }{ 0.35 }[/tex]
=> [tex]d = 9.04 *10^{-6 } \ m[/tex]
Experts in model airplanes develop a supersonic plane to scale, it moves horizontally in the air while it is conducting a flight test. The development team defines that the space that the airplane travels as a function of time is given by the function: e (t) = 9t 2 - 6t + 3 Determine what acceleration the scale airplane has (Second derivative).
Explanation:
e(t) = 9t² − 6t + 3
The velocity is the first derivative:
e'(t) = 18t − 6
The acceleration is the second derivative:
e"(t) = 18
The tires of a car make 77 revolutions as the car reduces its speed uniformly from 92.0 km/h to 60.0 km/h. The tires have a diameter of 0.84 m.
1. What was the angular acceleration of the tires?
2. If the car continues to decelerate at this rate, how much more time is required for it to stop?
3. If the car continues to decelerate at this rate, how far does it go? Find the total distance.
The angular acceleration of the tires is -2.2 rad/s².
If the car continues to decelerate at this rate, the time required to stop is 27.66 s.
The total distance traveled by the car before stopping is 210.96 revolutions.
The given parameters;
number of revolutions of the tire, N = 77 revinitial linear speed of the car, u = 92 km/h = 25.56 m/sfinal linear speed of the tire, v = 60 km/h = 16.67 m/sdiameter of the tire, d = 0.84 mradius of the tire, r = 0.42 mThe angular acceleration of the tire is calculated as follows;
[tex]\omega _f^2 = \omega _i ^2 + 2\alpha \theta\\\\(\frac{16.67}{0.42} )^2 = (\frac{25.56}{0.42} )^2 + 2( 77 \ rev \times \frac{2 \pi \ rad}{1 \ rev} ) \alpha \\\\1575.33 = 3703.59 \ + \ 967.736 \alpha \\\\-2128.26 = 967.736 \alpha\\\\\alpha = \frac{-2128.26}{967.736} \\\\\alpha = - 2.2 \ rad/s^2[/tex]
When the car stops, the final angular speed = 0. The time for the motion is calculated as;
[tex]\omega _f = \omega _i + \alpha t\\\\0 = \omega _i + \alpha t\\\\0 = 60.86 + (-2.2)t\\\\0 = 60.86 - 2.2t\\\\2.2t = 60.86\\\\t = \frac{60.86}{2.2} \\\\t = 27.66 \ s[/tex]
The total distance traveled by the car before stopping;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = (60.86 \times 27.66) \ + \ (0.5 \times -2.2\times 27.66^2)\\\\\theta = 841.8 \ rad\\\\\theta = 841.8 \ rad \times\frac{1 \ rev}{2\pi \ rad} = 133.96 \ rev[/tex]
total distance = 133.96 + 77 = 210.96 revolutions.
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