Answer:
A) Therefore laser1 has the maximum closest to the central maximum
B) Δₓ = 0.8
Explanation:
A) The expression for the constructive interference of a double slit is
d sin θ = m λ
let's use trigonometry to find the angle
tan θ = y / L
in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we subjugate
d y / L = m λ
y = m λ L / d
let's apply this equation for each case
a) Lares 1 has a wavelength λ₁ = d / 20, the screen is at L = 6.00 m
they ask us for the first axiom m = 1,
let's calculate
y₁ = 1 (d / 20) 6.00 / d
y₁ = 0.3
Laser 2, λ₂ = d / 15
λ₂ = 1 (d / 15) 6.00 / d
λ₂ = 0.4
Therefore laser1 has the maximum closest to the central maximum
b) let's find the distance of each requested value
second maximum m = 2 of laser 1
yi '= 2 (d / 20) 6 / d
y1 '= 0.6
3rd minimum of laser 2
the expression for destructive interference is
d sinθ = (m + 1/2) lam
y = (m ) λ L / d
in this case m = 3
let's calculate
y2 '= (3+0.5) (d / 15) 6 / d
y2 '=21/15
They ask us for the dalt of these interference
Δₓ = y3 -y2'
Δₓ = 21/15 - 0.6
Δₓ = 0.8
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]
If you converted 0.000013 to scientific notation, what would the prefix be to the correct number of significant digits?
Answer:
1.3
Explanation:
it will taken to as in from of standard form
Answer:
1.3 * 10^-5
Explanation:
We are learning about scientific notation.
When a number becomes a decimal followed before with zeroes, we know that the value of that number is decreasing. So instead of usually doing a positive exponent, we will do a negative exponent indicating we are going back.
So let's not only count the amount of zeroes followed before 13, but the decimal.
0.000013
The original number "1.3" went back 5 spaces, therefore making our exponent 5.
1.3 * 10^-5
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
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.
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
Turning the barrel of a 50-mm-focal-length lens on a manual-focus camera moves the lens closer to or farther from the sensor to focus on objects at different distances. The lens has a stated range of focus from 0.70 m infinity.
How far does the lens move between these two extremes?
Answer:
Explanation:
To focus object at .7m , the image distance can be measured as follows
object distance u = .7m
focal length f = .05 m
image distance v = ?
from lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{v} +\frac{1}{.7} = \frac{1}{.05}[/tex]
[tex]\frac{1}{v} =\frac{1}{.05} -\frac{1}{.7}[/tex]
v = .054 m
= 54 mm
when the object is at infinity , image is formed at focus ie at distance of
50 mm .
So lens position from sensor where image is formed , varies from 54 mm to 50 mm .
A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Answer:
After the bullet emerges the block moves at 0.99 m/s
Explanation:
Given;
mass of bullet, m₁ = 22 g = 0.022 kg
initial speed of the bullet, u₁ = 240 m/s
final speed of the bullet, v₁ = 150 m/s
mass of block, m₂ = 2.0 kg
initial speed of the block, u₂ = 0
Let the final speed of the block = v₂
Apply principles of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂
5.28 = 3.3 + 2v₂
5.28 - 3.3 = 2v₂
1.98 = 2v₂
v₂ = 1.98 / 2
v₂ = 0.99 m/s
Therefore, after the bullet emerges the block moves at 0.99 m/s
The body moves at a speed of 2.61m/s after the bullet emerges.
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
Based on the law above;
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
v is the final velocity of the body after the collision
Substitute the given parameters into the formula as shown:
[tex]0.022(240) + 2(0) = (0.022+2)v\\ 5.28 = 2.022v\\v=\frac{5.28}{2.022}\\v= 2.611m/s[/tex]
This shows that the body moves at a speed of 2.61m/s after the bullet emerges.
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You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 110 kg, perfectly absorbing block.
Required:
a. What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
b. Does this seem like a promising method for launching satellites?
Answer :
(a). The speed of the block is 0.395 m/s.
(b). No
Explanation :
Given that,
Diameter = 20.0 cm
Power = 26.0 MW
Mass = 110 kg
diameter = 20.0 cm
Distance = 100 m
We need to calculate the pressure due to laser
Using formula of pressure
[tex]P_{r}=\dfrac{I}{c}[/tex]
[tex]P_{r}=\dfrac{P}{Ac}
Put the value into the formula
[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}[/tex]
[tex]P_{r}=2.75\ N/m^2[/tex]
We need to calculate the force
Using formula of force
[tex]F=P\times A[/tex]
[tex]F=P\times \pi r^2[/tex]
Put the value into the formula
[tex]F=2.75\times\pi (0.01)^2[/tex]
[tex]F=0.086\ N[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F=ma[/tex]
Put the value into the formula
[tex]0.086=110\times a[/tex]
[tex]a=\dfrac{0.086}{110}[/tex]
[tex]a=0.000781\ m/s^2[/tex]
[tex]a=7.81\times10^{-4}\ m/s^2[/tex]
(a). We need to calculate speed of the block
Using equation of motion
[tex]v^2=u^2+2ad[/tex]
Put the value into the formula
[tex]v=\sqrt{2\times7.81\times10^{-4}\times100}[/tex]
[tex]v=0.395\ m/s[/tex]
(b). No because the velocity is very less.
Hence, (a). The speed of the block is 0.395 m/s.
(b). No
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]
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.
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
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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.
Zoning laws establish _______. a. what types of buildings can be built in an area b. the uses an area of land can be put to c. who can live in an area d. the types of business that can occupy a building Please select the best answer from the choices provided A B C D
Answer:
its B
Explanation:
Answer:
It's B
Explanation: hope it helps ^w^
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
sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A stationary observer hears a sound whose frequency is appreciably different from f because
Answer:
A static observer hears a sound whose frequency is appreciably different from actual frequency because here the change in frequency of the sound due to doppler effect.
Explanation:
Given that,
Frequency = f
We know that,
The sound source of frequency f moves with constant velocity through a medium that is at rest.
A static observer hears a sound whose frequency is significantly different from actual frequency due to doppler effect.
We know that,
The doppler effect is defined as
[tex]f=f_{0}(\dfrac{v+v_{0}}{v-v_{s}})[/tex]
Where, f₀ = actual frequency
f = observe frequency
v = speed of sound
[tex]v_{o}[/tex] = speed of observer
[tex]v_{s}[/tex] = speed of source
Hence, A static observer hears a sound whose frequency is significantly different from actual frequency because here the change in frequency of the sound due to doppler effect.
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]
#1 A boy pushes forward a cart of groceries with a total mass of 40 kg. What is
the acceleration of the cart if the net force on the cart is 60 N?
Explanation:
∑F = ma
60 N = (40 kg) a
a = 1.5 m/s²
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
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.
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]
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.
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
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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Consider the waveform expression. y(x,t)=ymsin(801t+3.38+0.503x) The transverse displacement ( y ) of a wave is given as a function of position ( x in meters) and time ( t in seconds) by the expression. Determine the wavelength, frequency, period, and phase constant of this waveform.
Answer:
f = 127.48 Hz , T = 7.844 1⁻³ s , Ф = 3.38 , λ = 12.49 m
Explanation:
The general equation for the motion of a wave in a string is
y = A sin (kx -wt + fi)
the expression they give is
y = ym sin (0.503x + 801 t + 3.38 )
the veloicda that accompanies time is
w = 801 rad / s
angular velocity is related to frequency
w = 2π f
f = w / 2π
f = 801 / 2π
f = 127.48 Hz
The period is the inverse of the frequency
T = 1 / f
T = 1 / 127.48
T = 7.844 10⁻³ s
the csntnate of phase fi is the independent term
Ф = 3.38
the wave vector accompanies the position k = 0.503 cm
ka = 2pi /λ
λ = 2 π / k
λ = 2 π / 0.503
λ = 12.49 m
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.
Air in a 124 km/h wind strikes head-on the face of a building 42 m wide by 73 m high and is brought to rest. If air has a mass of 1.3 kg per cubic centimeter, determine the average force of wind on the building.
Answer:
The average force of wind on the building is 4.728 x 10¹² N
Explanation:
Given;
speed of the air wind, v = 124 km/h
dimension of the building, 42 m wide by 73 m high
density of the air, ρ = 1.3 kg/cm³ =
speed of the air in m/s = 124/3.6 = 34.44 m/s
Area of the building, A = 42 m x 73 m = 3066 m²
density of the air in (k.g/m³);
[tex]\rho = \frac{1.3 \ kg}{cm^3} *(\frac{100\ cm}{1 \ m} )^3\\\\\rho = \frac{1.3 \ kg}{cm^3} *\frac{10^6\ cm^3}{1 \ m^3} = \frac{1.3*10^6 \ kg}{m^3}[/tex]
The average force of wind on the building;
F = mass flow rate x velocity
F = (ρvA) x V
F = ρAv²
F = 1.3 x 10⁶ x 3066 x (34.44)²
F = 4.728 x 10¹² N
Therefore, the average force of wind on the building is 4.728 x 10¹² N
Two parallel very long straight wires carrying current of 5A each are kept at a separation of 1m. If the currents are in the same direction, the force per unit length between them is __________
Answer:
The force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
Explanation:
Given;
current though the two parallel wires, I₁ and I₂ = 5A
distance between the two wires, R = 1 m
The force per unit of the wires is calculated as;
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Substitute in the given values into the equation and determine the force per unit length (F/L).
[tex]\frac{F}{L} = \frac{\mu_o I_1I_2}{2\pi R} \\\\ \frac{F}{L} = \frac{4\pi *10^{-7}*5*5}{2\pi *1}\\\\ \frac{F}{L} = 5*10^{-6} \ N/m \ (repulsive)[/tex]
Therefore , the force per unit between the two parallel wires with same current flowing in the same direction is 5 x 10⁻⁶ N/m repulsive force.
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
Please help and if you have the answer if you can please explain how you got it :)!
Answer:
The mass of ball C is greater than the mass of ball A but less than the mass of ball B.
Explanation:
From Newton's second law, net force = mass × acceleration.
Using the data for ball B, the acceleration of gravity near the surface of the moon is:
∑F = ma
9.6 N = (6 kg) a
a = 1.6 m/s²
Therefore, the mass of ball C is:
∑F = ma
6.6 N = m (1.6 m/s²)
m = 4.1 kg
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]