Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
What happens to the magnetic field when you reverse the direction of current by sliding the battery voltage bar past 0 volts
Answer:
The polarity of the magnetic field changes
Explanation:
This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)
g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
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What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.
Answer:
The thickness is [tex]t = 1.415 *10^{-7 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]
The index of refraction of glass is [tex]n_g = 1.71[/tex]
The index of refraction of the coating is [tex]n= 1.46[/tex]
Generally the condition for destructive interference is
[tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]
Here m is the order of the interference pattern and given from the question that we are considering minimizing reflection m = 0
t = thickness of the coating
substituting values
[tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]
=> [tex]t = 1.415 *10^{-7 } \ m[/tex]
Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.610 T.Required:a. What orientation of the coil gives the maximum torque on the coil ?b. What is this maximum torque in part (A) ?c. For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)?
Answer:
a) for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
b) therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
c) Given the torque is 71.0% of its maximum value; Ф = 45.24⁰ ≈ 45⁰
Explanation:
Given that; Diameter is 8.40 cm,
Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m
Number of turns (N) = 17
Current in the loop (I) = 3.20 A
Magnetic field (B) = 0.610 T
Let the angle between the loop's area vector A and the magnetic field B be
Now. the area of the loop is;
A = πR²
A = 3.14 ( 0.042 )²
A = 0.005539 m²
Torque on the loop (t) = NIABsinФ
t = 17 × 3.20 ×0.005539 × 0.610 × sinФ
t = 0.1838sinФ N.m
for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
Given the torque is 71.0% of its maximum value
t = 0.71 × tmax
t = 0.71 × 0.1838
t = 0.1305
Now
0.1305 N.m = 0.1838 sinФ N.m
sinФ = 0.1305 / 0.1838
sinФ = 0.71001
Ф = sin⁻¹ 0.71001
Ф = 45.24⁰ ≈ 45⁰
You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?
Answer:
Here,
v (final velocity) = 0
u (initial velocity) = u
a = ?
s = 1m
t = 0.4s
using the first equation of motion,
0 = u + 0.4a
= -0.4a = u
using the second equation of motion:
1 = 0.4u + 0.08a
from the bold equation
1 = 0.4(-0.4a) + 0.08a
1 = -0.16a + 0.08a
1 = -0.08a
a = -1/0.08
a = -100/8
a = -12.5 m/s/s
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What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm
Answer:
Explanation:
In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is
2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .
2 x 1.34 x t = 1 x 580
t = 216.42 nm .
Thickness must be 216.42 nm .
A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.
At what rate does this speaker produce energy?
What is the intensity of this sound 9.50 from the speaker?
What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
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In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.
Answer:
Explanation:
We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.
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A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?
Answer:
(a) nearsighted
(b) diverging
(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D
Explanation:
(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.
(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)
(c) the absolute value of the focal length (f) is given by the formula:
[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]
So it would normally be written with a negative signs in front indicating a divergent lens.
An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?
Answer:
200cm
Explanation:
Answer:
100cm
Explanation:
Using
F= ( N/2L)(√T/u)
F1 will now be (0.5*2)( √600/0.015)
=> L( wavelength)= 200/2cm = 100cm
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
Why was Bohr's atomic model replaced by the
modern atomic model?
Answer:
Explanation:
Bohr's atomic model was replaced by the modern atomic model because of its limitations, which included :
(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )
(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).
(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.
A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth
Answer:
t₀ = 1.55 s
Explanation:
According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in motion relative to the event is not the same as measured by an observer at rest.
It is given as:
t = t₀/[√(1 - v²/c²)]
where,
t = time measured by astronaut in motion = 3.1 s
t₀ = time required according to observer on earth = ?
v = relative velocity = 0.87 c
c = speed of light
3.1 s = t₀/[√(1 - 0.87²c²/c²)]
(3.1 s)(0.5) = t₀
t₀ = 1.55 s
Answer:
The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]Explanation:
Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]
where,
[tex]t_o = 3.1\\\\v = 0.87[/tex]
[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]
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You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.
Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.
Answer:
The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)Explanation:
WavelengthConvert all lengths to meters:
Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:
The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].
The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)
Therefore:
[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].
On the other hand:
[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].
Because the cotangent of [tex]\theta[/tex] is very close to zero,
[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].
[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].
[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].
Distance between two adjacent maximaIf the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.
The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:
[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].
Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction
Explanation:
In an electromagnetic wave the electric and magnetic fields are in phase
c = E / B
B = E / c
let's calculate
B = 4.00 / 3 10⁸
B = 1.33 10⁻⁸ T
To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.
If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction
A 26-g rifle bullet traveling 220 m/s embeds itself in a 3.8-kg pendulum hanging on a 2.7-m-long string, which makes the pendulum swing upward in an arc, Determine the vertical and horizontal component of the pendulum's maximum displacement
Answer:
displacements are 0.776m, 0.114m
Explanation:
We were given mass of 26-g rifle bullet , then we can convert to Kg since
Momentum is conserved here.
The initial momentum before impact = (Mi * Vi)
Where Mi= initial given mass
Vi=initial velocity given
= 0.026 * 220 = 5.72 kgm/s
The final momentum after impact is (Mf * Vf )
Mf= final mass
5.72=( 3.82* Vf )
= 5.72/ 3.82
= 1.497 m/s
the speed of the pendulum bob with bullet afterwards= 1.497 m/s
the total energy after the collision is the addition of the kinetic energy of the bob+bullet and the potential energy of the bob and bullet, potential energy can be taken as zero.
M = 3.82 kg the mass of the bob containing the bullet
E(total) = ¹/₂MV² = 1/2 * (3.82kg)*(1.497m/s)² = 4.280J
When the Bob got to highest point the kinetic energy is zero and the potential energy is due to the increase in height of the bob, and the addition of the potential and kinetic energies still equal the total energy from before
E(total) = Mgh + 0 = Mgh = 4.280J
solving for h and substituting,
h = 4.280 J/(9.8m/s^2*3.82kg) = 0.114 m
Since the height is found,we the angle of the pendulum at the top of the swing can also be determined
A = arccos[(2.7 - 0.114) / 2.7] or A = 16.71degrees
Since A is known, the displacement along the horizontal axis can be calculated as
x = 2.7* sin(A) = 0.776m
therefore, displacement is 0.776m, 0.114m
the vertical and horizontal component of the pendulum's maximum displacement are displacement is 0.776m, 0.114m
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the horizontal. The ball leaves her hand 1.0 m above the ground and the fence is 2.0 m high. The ball just clears the fence while still traveling upwards and experiences no significant air resistance. How far is the child from the fence?
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
[tex]x - x_o = u_xt[/tex]
[tex]\mathtt{x = u_xt \ \ \ since (x_o = 0)}[/tex] ---- (1)
the equation of the motion y is :
[tex]\mathtt{y - y_o =u_yt - 0.5 gt^2}[/tex]
[tex]\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}[/tex]
[tex]\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }[/tex]
[tex]\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}[/tex]
[tex]\mathtt{1 = 5.14t - 4.9t^2}[/tex]
[tex]\mathtt{4.9t^2 - 5.14t +1 = 0}[/tex]
By using the quadratic formula, we have;
[tex]\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }[/tex]
where;
a = 4.9, b = -5.14 c = 1
[tex]= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }[/tex]
[tex]= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }[/tex]
[tex]= \mathbf{ 0.791 \ \ OR \ \ 0.258} }[/tex]
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
[tex]\mathtt{x = u_x(0.258)}[/tex]
[tex]\mathtt{x = ucos 40^0 (0.258)}[/tex]
[tex]\mathtt{x = 8 \ cos 40^0 (0.258)}[/tex]
[tex]\mathbf{x = 1.581 \ m}[/tex]
Thus, the child is 1.581 m far from the fence
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).
Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answer:
a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².
Explanation:
a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.
τ = Frsinθ
= 185 N × 1.73 m × sin90°
= 320.05 Nm
So, the torque τ = 320.05 Nm
b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m
α = angular acceleration of grindstone
τ = Iα
α = τ/I = τ/(MR²/2) = 2τ/MR²
substituting the values of the variables, we have
α = 2τ/MR²
= 2 × 320.05 Nm/[76 kg × (1.73 m)²]
= 640.1 Nm/227.4604 kgm²
= 2.814 rad/s²
So, the angular acceleration α = 2.814 rad/s²
c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.
So τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm
The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm
Since τ'' = Iα
α' = τ''/I where α' = its new angular acceleration
α' = 2τ/MR²
= 2 × 319.75 Nm/[76 kg × (1.73 m)²]
= 639.5 Nm/227.4604 kgm²
= 2.811 rad/s²
So, the angular acceleration α' = 2.811 rad/s²
