Answer:
a) 15.864s
b) 392.78m
c) 29.52 m/s
Explanation:
The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is
distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)
= 24 + 21 + 26 + 4.5 = 75.5 m
As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck
[tex]s = at^2/2[/tex]
[tex]75.5 = 0.6t^2/2[/tex]
[tex]t^2 = 251.67[/tex]
[tex]t = \sqrt{251.67} = 15.864s[/tex]
b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time
= 75.5 + 20*15.864 = 392.78 m
c) final speed of the car is the initial speed plus the change in speed
[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]
To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.
Explanation:u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.
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A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the metal (M). Please provide your answer in 2 decimal places.
Answer: The molar mass of metal (M) is 47.86 g/mol
Explanation:
Let the atomic mass of metal (M) be x
Atomic mass of [tex](MO_2)=(x+32)g/mol[/tex]
To calculate the mass of metal, we use the equation:
[tex]\text{Mass percent of metal}=\frac{\text{Mass of metal}}{\text{Mass of metal oxide}}\times 100[/tex]
Mass percent of metal = 59.93 %
Mass of metal = x g/mol
Mass of metal oxide = (x + 32) g/mol
Putting values in above equation, we get:
[tex]59.93=\frac{x}{(x+32)}\times 100\\\\59.93(x+32)=100x\\\\x=47.86g/mol[/tex]
Hence, the molar mass of metal (M) is 47.86 g/mol
A single water molecule were oriented such that its dipole moment (magnitude 6.186 × 10 − 30 Cm is along a z axis. How much torque does an 8500 N/C electric field exert on the dipole if the field lies in x z -plane, 42 ∘ above the x -axis?
Answer:
Torque=6.261×10^-43Nm
Explanation:
Torque is given by÷
Torque=dqEsin(phi)
Where E is a constant=1.6022×10^-19
d= distance = 6.186×10^-30cm
Changing to metres=6.186×10^-28m
q= the charge=8500NC
Phi =42°
Torque= (6.186×10^-28)×8500×(1.6022×10^-19)sin42°
Torque= 8.425×10^-43 × 0.7431
Torque= 6.261 ×10^-43
The torque exerted by an 8500 N/C electric field exerts on the dipole is 6.261 ×10⁻⁴³ Nm.
What is torque?The force which causes the object to rotate about any axis is called perpendicular distance.
Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.
Given:
A single water molecule was oriented such that its dipole moment of magnitude 6.186 × 10 − 30 Cm is along a z-axis,
The electric field = 8500 N/C,
∅ = 42°
Calculate the torque as shown below,
Torque = d × q × E × sin(∅)
Where E is a constant = 1.6022×10⁻¹⁹
distance = 6.186×10⁻³⁰ cm = 6.186×10⁻²⁸ m
Torque = 6.186×10⁻²⁸ × 8500 × (1.6022 × 10⁻¹⁹) sin42°
Torque = 8.425×10⁻⁴³ × 0.7431
Torque = 6.261 ×10⁻⁴³
Thus, the torque is 6.261 ×10⁻⁴³ Nm.
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An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of q on the plane and a point charge of -q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 53.5 J. Find the magnitude of the charges.
Answer:
q=3.5*10^-4
Explanation:
concept:
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
The tension force:
when the plane is uncharged
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
when the plane is charged
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
A car of mass m push = 1200 kg is capable of a maximum acceleration of 6.00 m / s 2 . If this car is required to push a stalled car of mass m stall = 1750 kg, what is the maximum magnitude of the acceleration a of the two‑car system? a = 0.62 m / s 2
1)
first you find the maxium force that the car can produce.
f=ma
Fmax=(1100kg)(6m/s^2)
then use f = ma again to find the accel with the passengers
Fmax=(1100kg +1650kg)(a)
=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)
= 2.4 m/s^2
The maximum magnitude of the acceleration of the two‑car system is [tex]2.44 \;\rm m/s^{2}[/tex].
Given data:
The mass of car is, m = 1200 kg.
The maximum acceleration is, [tex]a =6.00 \;\rm m/s^{2}[/tex].
The mass of stall is, m' = 1750 kg.
As per the Newton's Second law of motion, the applied force on the car is equal to the product of mass and acceleration of car. So, the maximum force to push the car is given as,
F = ma
[tex]F = 1200 \times 6.00\\\\F= 7200\;\rm N[/tex]
Now for two-car system, the total mass is,
M = m + m'
M = 1200 +1750 = 2950 kg.
So, the acceleration for the two-car system is,
[tex]F = M \times a'\\\\7200 =2950 \times a'\\\\a'=2.44 \;\rm m/s^{2}[/tex]
Thus, the maximum magnitude of the acceleration a of the two‑car system is [tex]2.44 \;\rm m/s^{2}[/tex].
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Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4+????2.7) Ω per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2200√3 V. Compute (a) the line-to-line voltage magnitude (in RMS value) at the source end of the line, (b) the total real and reactive power losses in the three-phase line, and (c) the total real and reactive power supplied at the sending end of the line.
Answer:
find answers below
Explanation:
a)
S1 560 e^(j acos 0.707 ) ⋅kVA S1 = ( ) 395.92 396.04j kW+ ⋅
S2 := 132 kW⋅ S2 = 132 kW⋅
Sd S1 +S2 :=
Sd = 527.92 396.04j kW
Sd = 660 kVA ⋅
arg Sd = 36.877 deg ⋅
Vd 2.2 e^(− j⋅0⋅deg)⋅kV
current in the line =(Sd/3 Vd) ⋅
⎯
:=
I Line = 79.988 60.006j A
Line = 99.994 A
arg I( ) Line = −36.877⋅deg
r Line := 0.4⋅Ω resistance in the line xLine := 2.7⋅Ω
Vsan Vd+ (r Line+ j xLine) ILine
Vsan = ( ) kV 2.394 0.192j + ⋅ Vsan = 2.402 kV⋅ arg V( ) san = 4.584 deg ⋅
Vsab 3 e
j 30 ⋅ ⋅deg ⋅ Vsan := ⋅
Vsab = ( ) kV 3.425 2.361j + ⋅ Vsab = 4.16 kV⋅ arg V( ) sab = 34.584 deg
b)
PLine 3 I ( ) Line
2 ⋅ r
Line := ⋅
PLine = 12 kW⋅
QLine 3 I ( ) Line
2 ⋅ xLine := ⋅ QLine = 80.99 kVAR ⋅
c)
Ss 3 Vsan ⋅ I
Line := ⋅
⎯ Ss = ( ) kVA 539.919 477.03j + ⋅
Ss = 720.5 kVA ⋅ arg S( )s = 41.461 deg ⋅
Ps Re S( )s := Ps = 540 kW⋅
Qs Im S( )s := Qs = 477 kVAR ⋅
S1 S2 + PLine + j QLine + ⋅ = ( ) kVA 539.919 477.03j + ⋅ Check
Suppose a hydrogen atom in its ground state moves 130 cm through and perpendicular to a vertical magnetic field that has a magnetic field gradient dB/dz = 1.2 × 102 T/m. (a) What is the magnitude of the force exerted by the field gradient on the atom due to the magnetic moment of the atom's electron, which we take to be 1 Bohr magneton? (b) What is the vertical displacement of the atom in the 130 cm of travel if its speed is 2.2 × 105 m/s?
Answer:a)1.11×10^-21Nm
b) 1.16×10^-3m
Explanation:see attachment
The energy of a photon is inversely proportional to the wavelength of the radiation. (T/F)
Answer:
true
Explanation:
The statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
What is Wavelength?Wavelength may be characterized as the space or distance between two successive crests of a wave that especially includes the points in a sound wave or electromagnetic wave. The wavelength may be represented by a letter known as lambda (λ).
When the energy of the radiation increases, the waves travel faster which directs the small gap between two successive crests. This means that the wavelength decreases.
On contrary, when the energy of the radiation decreases, the waves travel slower which leads to a huge gap between two successive crests. This means that the wavelength increases.
Therefore, the statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
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An X-ray telescope located in Antarctica would not work well because of (a) the extreme cold; (b) the ozone hole; (c) continuous daylight; (d) Earth’s atmosphere.
The entrance of certain frequencies through the atmosphere varies depending on the sector. This concept of permittivity by which the rays can enter the earth is called the Radio Window and has a spectrum that ranges from 5MHz to 30GHz. Of the radiation produced by astronomical objects, one part is 'filtered' and the other has the ability to be absorbed by the earth due to its opacity effect.
Ultraviolet rays, X-rays and even gamma rays are completely affected by the ozone layer in Earth's atmosphere.
This makes it difficult to use X-ray telescopes located in Antarctica.
The correct option is D.
Earth's atmosphere limits X-ray observations, making Antarctica unsuitable due to extreme cold.
Earth's atmosphere acts as a barrier for X-ray observations, as it blocks most radiation at wavelengths shorter than visible light. Telescopes located in Antarctica would not work well due to the extreme cold which can affect their performance.
A bullet moving with an initial speed of vo strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true of the collision? o The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. the bullet immediately after the collision the potential energy of the bullet and block at the instant they reach the maximum O The initial momentum of the bullet before the collision is equal to the momentum of o The kinetic energy of the bullet and block immediately after the collision is equal to height h. energy of the bullet and block when they reach the maximum heighth energy of the bullet and block immediately after the colision O The initial kinetic energy of the bullet before the collision is equal to the potential o The initial kinetic energy of the bullet before the collision is equal to the kinetic
Final answer:
The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. Option A
Explanation:
The correct statement of the collision in this scenario is A) The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. In this case, the bullet and block form a closed system, and therefore, the initial momentum of the bullet is equal to the momentum of the bullet and block at the maximum height.
To understand why this statement is true, we can consider the momentum of the bullet and block system at different points in the motion. Initially, the bullet has momentum in the positive x-direction, and this momentum is transferred to the bullet and block as they rise to the maximum height. Therefore, the statement A is correct.
Your electricity bill says 834-kWh of consumption. With this amount of energy, how many 59-W light bulbs can be powered for an average of 7 hours?
Answer:
2019 light bulbs
Explanation:
So the electrical energy consumed by each 59-W light bulb within 7 hours is the product of the power and time duration
E = Pt = 59 * 7 = 413 Wh or 0.413 kWh
If each light bulb consumes 0.413 kWh, then the total number of light bulbs needed to consume 834 kWh would be
834 / 0.413 = 2019 light bulbs
Water on Earth was (a) transported here by comets; (b) accreted from the solar nebula; (c) produced by volcanoes in the form of steam; (d) created by chemical reactions involving hydrogen and oxygen shortly after Earth formed.
Answer: Water on Earth was transported here by comets. The correct option is A.
Explanation:
Comets are made up of water with ice, rock and minerals.
Alot of research and hypotheses has been made to prove the origin of water on planet earth. Extraplanetary source such as comets, trans-Neptunian objects, and water-rich meteoroids (protoplanets) are believed to have delivered water to Earth.
Final answer:
Water on Earth primarily originated from the (b) solar nebula during planetary accretion, with minor contributions from comets and volcanic activity. Hence, (b) is the correct option.
Explanation:
As Earth coalesced from the protoplanetary disk around the young Sun, volatile compounds, including water, were incorporated into the developing planet. Comets may have contributed some water, but the predominant source was the solar nebula.
While volcanic activity released water vapor, it played a minor role compared to the water acquired during accretion. Chemical reactions involving hydrogen and oxygen likely occurred after Earth's formation, but the bulk of the water was established during the early stages of planetary accretion from the solar nebula.
Normal human body temperature is about 37°C. What is this temperature in kelvins? What is the peak wavelength emitted by a person with this temperature? In what part of the spectrum does this lie?
Normal human body temperature is about 37°C, which is equivalent to 310.15 K. The peak wavelength emitted by a person with this temperature is 9.35 μm, which lies in the infrared part of the spectrum.
Explanation:Normal human body temperature is about 37°C. To convert this temperature to kelvin, we can use the formula K = °C + 273.15. So, the temperature in kelvins would be:
K = 37 + 273.15 = 310.15 K.
The wavelength emitted by a person with this temperature can be determined using Wien's displacement law: λ = b/T, where b is Wien's constant and T is the absolute temperature. The peak wavelength (λmax) is the wavelength at which the emitted radiation is the greatest. For a temperature of 310.15 K, the peak wavelength can be calculated as: λmax = b/T = 2897.8/T = 9.35 μm.
The peak wavelength of 9.35 μm corresponds to the infrared part of the electromagnetic spectrum.
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Whose view was that the Cosmos was based on the belief that every occurrence in the physical universe had logos behind it and that is where life originated?
Answer:
Aristotle
Explanation:
Aristotelian theory of the Universe
For two millennia, the philosophical tradition considered that the universe was eternal and did not change. The wise Aristotle said so, with total clarity and his ideas dominated Western thought for more than two thousand years.
This distinguished philosopher believed that the stars are made of an imperishable matter and that the landscapes of the sky are immutable.
From the time of Aristotle until the beginning of the twentieth century, the idea that the universe was static, that the cosmos had been eternally equal, was admitted.
In those years, the origin of the universe was not really considered in a scientific way, since it was based on the basis that the gods had created everything that exists, at the time they wanted it, according to their omnimous power.
So that all the efforts of the wise men of the time focused on discovering the existing organization in the universe created by the gods.
According to Aristotle and the thinkers of the fourth century B.C. what is below the Moon is a changing world, what is beyond the Moon is an immutable world.
A 5.55 L cylinder contains 1.21 mol of gas A and 4.93 mol of gas B, at a temperature of 27.3 °C. Calculate the partial pressure of each gas in the cylinder. Assume ideal gas behavior.
Answer:
Pressure of gas A = 544.324 kPa
Pressure of gas B = 2217.784 kPa
Explanation:
Data provided in the question:
Total volume of the cylinder, V = 5.55 L = 0.00555 m³ [1 m³ = 1000 L]
Moles on gas A, [tex]n_a[/tex] = 1.21 mol
Moles on gas A, [tex]n_b[/tex] = 4.93 mol
Temperature, T = 27.3°C = 27.3 + 273 = 300.3 K
Now,
Pressure = [tex]\frac{nRT}{V}[/tex]
here,
R is the ideal gas constant = 8.314 J/mol.K
Therefore,
Pressure of gas A = [tex]\frac{n_aRT}{V}[/tex]
= [tex]\frac{1.21\times8.314\times300.3}{0.00555}[/tex]
= 544324.32 Pa
= 544.324 kPa
Pressure of gas B = [tex]\frac{n_bRT}{V}[/tex]
= [tex]\frac{4.93\times8.314\times300.3}{0.00555}[/tex]
= 2217784.22 Pa
= 2217.784 kPa
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 kg of turkey. The slices of turkey are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of turkey are dropped on the plate all at the same time from a height of 0.250 m . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.a. What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?b. What is the period of oscillation T of the scale?
Answer:
a) A = 0.0221 m, b) T = 0.314 s
Explanation:
For this exercise we must separate the process into two parts, one when the body falls and another when it hits the tray
Let's look for the speed with which it reaches the tray, using energy conservation
Initial. Highest point
Em₀ = U = m g h
Final. Tray Point
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v = √2gh
Now let's apply moment conservation
Initial. Just before the crash
p₀ = m v
Final. After the crash
p_{f} = (m + M) v_{f}
p₀ = p_{f}
m v = (m + M) v_{f}
v_{f} = m / (m + M) √2gh
This is the turkey + plate system speed, which is the one that will oscillate
b) The angular velocity of the oscillation is
w = √ k / m
The angular velocity is related to the frequency and period
w = 2πf = 2π / T
T = 2π √m / k
T = 2π √ ((0.100 + 0.400) / 200)
T = 0.314 s
a) To find the amplitude let's use the equation that describes the oscillatory motion
y = A cos (wt + fi)
Speed is
v = dy / dt = - A w sin (wt + fi)
At the initial point the turkey + plate system has a maximum speed, in the previous equation the speed is maximum when cos (wt + fi) = ±1
v = v_{f} = A w
m / (m + M) √ 2gh = A w
A = m / (m + M) √2gh 1 / w
w = 2π / T
Let's calculate
A = 0.100 / (0.100 + 0.400) √(2 9.8 0.250) 0.314/2π
A = 0.2 2.2136 0.049975
A = 0.0221 m
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 95.0mi/h fastball headed straight toward the gun?
Since the Units presented are not in the International System we will proceed to convert them. We know that,
[tex]1 mi/h = 0.447 m/s[/tex]
So the speed in SI would be
[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]
[tex]V=42.465 m/s[/tex]
The change in frequency when the wave is reflected is
[tex]f'=f(1+\frac{V}{c})[/tex]
Or we can rearrange the equation as
[tex]f' = f + f\frac{V}{c}[/tex]
f' = Apparent frequency
f = Original Frequency
c = Speed of light
[tex]f'-f = f\frac{V}{c}[/tex]
[tex]\Delta f = f\frac{V}{c}[/tex]
Replacing,
[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]
[tex]\Delta f =1489.8 Hz[/tex]
Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency
[tex]\Delta f_T = 2 \Delta f[/tex]
[tex]\Delta f_T = 2(1489.8Hz)[/tex]
[tex]\Delta f_T = 2979.63Hz[/tex]
Therefore the increase in frequency is 2979.63Hz
The increase in frequency of these waves is equal to 2979.6 Hertz.
Given the following data:
Frequency = 10.525 GHz.Velocity = 95.0 mi/h.Conversion:
1 mi/h = 0.447 m/s.
95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.
To determine the increase in frequency:
How to calculate the increase in frequency.Mathematically, the change in frequency of a wave is given by this formula:
[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]
Where:
F is the observed frequency.[tex]F'[/tex] is the apparent frequency.c is the speed of light.V is the velocity of an object.Substituting the given parameters into the formula, we have;
[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]
Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;
[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]
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A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b) What is the radial acceleration of the blade tip expressed as a multiple of g?
the radial acceleration of the blade tip is approximately 1210.50 times g .
To find the linear speed of the blade tip and the radial acceleration of the blade tip, we can use the following formulas:
(a) Linear speed of the blade tip:
[tex]\[ \text{Linear speed} = \text{Angular speed} \times \text{Radius} \]\[ v = \omega \times r \][/tex]
(b) Radial acceleration of the blade tip:
[tex]\[ \text{Radial acceleration} = \omega^2 \times r \]\[ a_r = \omega^2 \times r \][/tex]
Given:
- Angular speed [tex](\( \omega \))[/tex] = 550 rev/min
- Radius [tex](\( r \))[/tex] = 3.40 m
- Acceleration due to gravity[tex](\( g \))[/tex] = 9.81 m/s²
First, let's convert the angular speed from rev/min to rad/s:
[tex]\[ \omega = \frac{550 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ s/min}} \]\[ \omega = \frac{550 \times 2\pi}{60} \text{ rad/s} \]\[ \omega \approx 57.91 \text{ rad/s} \][/tex]
(a) Linear speed of the blade tip:
[tex]\[ v = \omega \times r \]\[ v = 57.91 \times 3.40 \]\[ v \approx 197.04 \text{ m/s} \][/tex]
So, the linear speed of the blade tip is approximately 197.04 m/s.
(b) Radial acceleration of the blade tip:
[tex]\[ a_r = \omega^2 \times r \]\[ a_r = (57.91)^2 \times 3.40 \]\[ a_r \approx 11854.76 \text{ m/s}^2 \][/tex]
Now, express the radial acceleration as a multiple of ( g ):
[tex]\[ \text{Multiple of } g = \frac{a_r}{g} \]\[ \text{Multiple of } g = \frac{11854.76}{9.81} \]\[ \text{Multiple of } g \approx 1210.50 \][/tex]
So, the radial acceleration of the blade tip is approximately 1210.50 times ( g ).
Two small plastic spheres each have a mass of 1.2 g and a charge of -56.0 nC . They are placed 3.0 cm apart (center to center). Part A What is the magnitude of the electric force on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
F_e = +/- 0.013133 N
Explanation:
Given:
- charge on each sphere q = -56 nC
- separation of spheres r = 3.0 cm
- charge constant k = 8.99*10^9
Find:
- Magnitude of Electric Force F_e on each sphere.
Solution:
The magnitude of electric Force F_e of one sphere on other is:
F_e = k*q^2 / r^2
- Plugging the given values :
F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2
F_e = 0.013133 N
- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N
Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant.A.What is the x component Fx of the resultant force?Express your answer in newtons.B. What is the y component Fy of the resultant force?Express your answer in newtons.C. What is the magnitude F of the resultant force?Express your answer in newtons.D. What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.Express your answer in degrees.
A) 7.297 N is the x component Fx of the resultant force.
Part B) - 3.654 N is the y component Fy of the resultant force.
Part C) 8.16N is the magnitude F of the resultant force.
D) 26.55° from the negative x-axis in a clockwise direction.
A) x-component of forces is 7.297 N
For x-component FX;
FX = F1 cos 65 + F2 cos 53.9
FX = 9.20 cos 65 + 5.8 cos 53.9
NB both forces are positive if resolved along the x-axis.
FX = 3.88 + 3.417 = 7.297 N
B) 6-component of forces is -3.654 N
For y-component FY
FY = -F1 sin 65 + F2 sin 53. 9
FY = - 9.2 sin 65 + 5.8 sin 53.9
NB F1 is negative along y-axis
FY = - 8.34 + 4.686 = - 3.654 N
C) The magnitude of the resultant force is 8.16 N
Magnitude :
[tex](FX^2 + FY^2)^{0.5}\\(7.297^2 + (-3.654)^2)^{0.5} \\= 8.16 \text N[/tex]
D) Angle made with negative x-axis is 26.56°
[tex]\text{Tan}^{ -1[/tex] of FY/FX gives the angle of the resultant force.
= [tex]\text{tan}^{ -1[/tex] of -3.654/7.297
=[tex]\text{tan}^{ -1[/tex] of -0.5
= -26.56°
i.e. 26.55° from the negative x-axis in a clockwise direction.
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(A) The x-component of the resultant force is approximately 7.039 N.
(B) The y-component of the resultant force is approximately 3.654 N.
(C) The magnitude of the resultant force is approximately 7.95 N.
(D) The angle θ with the negative x-axis is approximately 28.6 degrees (measured clockwise).
To find the x and y components of the resultant force, as well as its magnitude and angle with the negative x-axis, we can use vector addition. Given the forces F₁ and F₂:
F₁ = 9.20 N (65.0° above the negative x-axis)
F₂ = 5.80 N (53.9° below the negative x-axis)
(A) Finding the x-component (Fₓ) of the resultant force:
We'll first find the x-component of each force:
For F₁:
F₁ₓ = F₁ * cos(65.0°)
For F₂:
F₂ₓ = F₂ * cos(53.9°)
Now, calculate the net x-component:
Fₓ = F₁ₓ + F₂ₓ
Calculate Fₓ:
F₁ₓ = 9.20 N * cos(65.0°) ≈ 4.016 N
F₂ₓ = 5.80 N * cos(53.9°) ≈ 3.023 N
Fₓ = 4.016 N + 3.023 N ≈ 7.039 N
(B) Finding the y-component (Fᵧ) of the resultant force:
Similarly, we'll find the y-component of each force:
For F₁:
F₁ᵧ = F₁ * sin(65.0°)
For F₂:
F₂ᵧ = -F₂ * sin(53.9°) [negative because it's below the x-axis]
Now, calculate the net y-component:
Fᵧ = F₁ᵧ + F₂ᵧ
Calculate Fᵧ:
F₁ᵧ = 9.20 N * sin(65.0°) ≈ 8.111 N
F₂ᵧ = -5.80 N * sin(53.9°) ≈ -4.457 N
Fᵧ = 8.111 N - 4.457 N ≈ 3.654 N
(C) Finding the magnitude (F) of the resultant force:
The magnitude of the resultant force is given by the Pythagorean theorem:
F = √(Fₓ² + Fᵧ²)
Calculate F:
F = √(Fₓ² + Fᵧ²) ≈ √(7.039 N)² + (3.654 N)² ≈ √(49.53 N² + 13.35 N²) ≈ √62.88 N² ≈ 7.95 N
(D) Finding the angle (θ) with the negative x-axis:
The angle θ with the negative x-axis can be found using the inverse tangent function:
θ = arctan(Fᵧ / Fₓ)
Calculate θ:
θ = arctan(Fᵧ / Fₓ) ≈ arctan(3.654 N / 7.039 N) ≈ arctan(0.519) ≈ 28.6°
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What is the Doppler effect, and how does it alter the way in which we perceive radiation?
Answer:
Explanation:
When the distance is decreasing, the frequency of the received wave form will be higher than the source wave form. Besides sound and radio waves, the Doppler effect also affects the light emitted by other bodies in space. If a body in space is "blue shifted," its light waves are compacted and it is coming towards us.
The Doppler effect is a phenomenon that describes how the perceived frequency and wavelength of waves change due to relative motion. It can result in a blue shift or a red shift, depending on the direction of motion. The Doppler effect has applications in astronomy and medicine, among other fields.
Explanation:The Doppler effect is a phenomenon in physics that describes how the perceived frequency and wavelength of waves change when there is relative motion between the source of the waves and the observer. This effect can alter the way in which we perceive radiation, such as light or sound.
When an object emitting waves moves towards an observer, the waves are compressed, resulting in a higher frequency and shorter wavelength. This is known as a blue shift. On the other hand, when an object moves away from an observer, the waves are stretched, resulting in a lower frequency and longer wavelength, which is called a red shift.
The Doppler effect has several applications in various fields. For example, it is used in astronomy to determine the speed and direction of celestial objects based on their red or blue shifts. In medicine, it is used in ultrasound technology to measure blood flow velocity. Overall, the Doppler effect allows us to understand and interpret the motion of objects emitting waves.
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Two equal positive point charges are placed at two of the co ?rners of an equilateral triangle of side A. What is the magnitude of the net electric field at the center of the triangle ?
Answer:
Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C
Explanation:
Given,A = side of the triangle. q = 1 C
The center of a triangle is the centroid of the triangle.
The distance between centroid to any vertices of a equilateral triangle is
[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle
[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]
[tex]=\frac{1}{\sqrt{3} } A[/tex]
Electric field= [tex]\frac{Kq}{d^2}[/tex] K=8.99×10⁹ Nm²/C², q=charge and d = distance
Therefore the magnitude of the net electric filed at the center of the triangle is
=2[tex]\frac{Kq}{d^2}[/tex] [both charge are at same distance from the centroid]
=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]
[tex]=\frac{6K}{A^2}[/tex] N /C [K=8.99×10⁹ Nm²/C²]
A rigid tank contains air at a pressure of 70 psia and a temperature of 55 ˚F. By how much will the pressure increase as the temperature is increased to 115 ˚F?
Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,
[tex]\dfrac{P_2}{P_1} =\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{P_2}{70} =\dfrac{55+459.67}{115+459.67}[/tex]
P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi
Perform the following unit conversions:
a. 1 L to in.3
b. 650 J to Btu
c. 0.135 kW to ft · lbf/s
d. 378 g/s to lb/min
e. 304 kPa to lbf/in.2
f. 55 m3/h to ft3/s
g. 50 km/h to ft/s
h. 8896 N to ton (=2000 lbf)
Explanation:
a. 1 liter (L) is equal to 61.0237 cubic inches ([tex]in^3[/tex]), So:
[tex]1L*\frac{61.0237in^3}{1L}=61.0237in^3[/tex]
b. 1 J is equal to [tex]9.47817*10^{-4}[/tex] BTU. Thus:
[tex]650J*\frac{9.47817*10^{-4}BTU}{1J}=6.16081*10^{-1}BTU[/tex]
c. 1 kW is equal to 737.56 [tex]\frac{ft\cdot lbf}{s}[/tex]. So:
[tex]0.135kW*\frac{737.56\frac{ft\cdot lbf}{s}}{1kW}=99.57\frac{ft\cdot lbf}{s}[/tex]
d. 1 [tex]\frac{g}{s}[/tex] is equal to 0.1323 [tex]\frac{lb}{min}[/tex]. Therefore:
[tex]378\frac{g}{s}*\frac{0.1323\frac{lb}{min}}{1\frac{g}{s}}=50.01\frac{lb}{min}[/tex]
e. 1 kPa is equal to 0.145[tex]\frac{lbf}{in^2}[/tex]. Thus:
[tex]304kPa*\frac{0.145\frac{lbf}{in^2}}{1kPa}=44.08\frac{lbf}{in^2}[/tex]
f. 1 [tex]\frac{m^3}{h}[/tex] is equal to [tex]9.81*10^{-3}\frac{ft^3}{s}[/tex]. So:
[tex]55\frac{m^3}{h}*\frac{9.81*10^{-3}\frac{ft^3}{s}}{1\frac{m^3}{h}}=5.40*10^{-1}\frac{ft^3}{s}[/tex]
g. 1 [tex]\frac{km}{h}[/tex] is equal to 0.911 [tex]\frac{ft}{s}[/tex]. Therefore:
[tex]50\frac{km}{h}*\frac{0.911\frac{ft}{s}}{1\frac{km}{h}}=45.55\frac{ft}{s}[/tex]
h. 1 N is equal to [tex]1.1*10^{-4}[/tex] ton. Thus:
[tex]8896N*\frac{1.1*10^{-4}ton}{1N}=0.98ton[/tex]
Two balls of different radii, 53 cm and 26 cm move directly toward each other with the same speed. If they are originally at a center-to-center distance 223 cm and it takes them 18.9 s to collide, how fast were the balls moving?
Answer:
[tex]v = 3.81\ m/s[/tex]
Explanation:
given,
Radius of the ball, r₁ = 53 cm
Radius of another ball, r₂ = 26 cm
center to center distance between the balls = 223 cm
time, t = 18.9 s
surface to surface distance between them
S = 223 - (53+26)
S = 144 cm
Speed of the ball = ?
[tex]relative\ speed = \dfrac{distance}{time}[/tex]
[tex]2 v = \dfrac{144}{18.9}[/tex]
both the balls are moving towards each other so, speed doubles.
[tex]v= \dfrac{7.62}{2}[/tex]
[tex]v = 3.81\ m/s[/tex]
Speed of the balls is equal to 3.81 m/s
A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0 s, what is the current in the wire? 4.5 A 18 A 0.22 A 9.0 A If the current is a constant 3.0 A, how long will it take for 14.0 C of charge to move past a particular spot in the wire? 0.21 s 4.7 s 14.0 s 42 s
1) Current: 4.5 A
2) Time taken: 4.7 s
Explanation:
1)
The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
For the wire in this problem, we have
q = 9.0 C is the amount of charge
t = 2.0 s is the time interval
Solving for I, we find the current:
[tex]I=\frac{9.0}{2.0}=4.5 A[/tex]
2)
To solve this problem, we can use again the same formula
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
In this problem, we have:
I = 3.0 A (current)
q = 14.0 C (charge)
Therefore, the time taken for the charge to move past a particular spot in the wire is
[tex]t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s[/tex]
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The electric current in the wire in the first scenario is 4.5 A, and it will take 4.7 seconds for 14.0 C of charge to move past a particular spot on the wire with a constant current of 3.0 A in the second scenario.
Explanation:The amount of electric current in a circuit is defined by the amount of electric charge that passes through it in a given amount of time. This is represented by the formula I = Q / t, where I is the current, Q is the charge and t is the time.
In the first part of your question, we are given that Q = 9.0 C and t = 2.0 s. We can substitute these values into the formula to find the current: I = 9.0 C / 2.0 s = 4.5 A.
In the second part, we are given that I = 3.0 A and Q = 14.0 C. This time, we need to rearrange the formula to find t: t = Q / I = 14.0 C / 3.0 A = 4.7 s.
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Consider a horizontal semi-transparent plate at a temperature 360 K that is irradiated from above and below. Air at [infinity] 310 K flows over both sides of the plate with a convective coefficient, 45 W/m2·K. The absorptivity of the plate is 0.4. If the radiosity is 4500 W/m2, find the irradiation, in W/m2, and emissivity of the plate. Is the plate diffuse-gray?
Answer:
[tex]G=6750\ W.m^2[/tex] is the irradiation
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
Explanation:
Given that:
Temperature of air, [tex]T_{\infty}=310\ K[/tex]
temperature of plate, [tex]T_s=360\ K[/tex]
convective heat transfer coefficient, [tex]h=45\ W.m^{-2}.K^{-1}[/tex]
absorptivity of plate, [tex]\alpha=0.4[/tex]
radiosity of the plate, [tex]J=4500\ W.m^{-2}[/tex]
From the energy balance eq. :
[tex]E_{in}=E_{out}[/tex]
since two surfaces are involved
[tex]2G=2J+2q_{conv}[/tex]
[tex]G=J+q_{conv}[/tex]
where
[tex]q_{conv}=[/tex] heat transfer per unit area due to convection
[tex]G=[/tex] irradiation per unit area
[tex]G=J+h.\Delta T[/tex]
[tex]G=4500+45\times (360-310)[/tex]
[tex]G=6750\ W.m^2[/tex] is the irradiation
Since radiosity includes transmissivity, reflectivity and emissivity.
[tex]J=G.\tau+G.\rho+E[/tex]
[tex]J=G.\tau+G.\rho+\epsilon\times \sigma\times T^{4}[/tex]
where:
[tex]\tau=[/tex] transmissivity
[tex]\rho=[/tex] reflectivity
[tex]\epsilon=[/tex] emissivity
Since the absorptivity of the plate is 0.4 and the plate is semitransparent so the transmissivity and reflectivity = 0.3 each.
[tex]4500=0.3\times 6750+0.3\times 6750+\epsilon\times 5.67\times 10^{-8}\times 360^4[/tex]
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
The Stefan-Boltzmann law of radiation can be used to find the irradiation and emissivity of a plate. The irradiation can be found using the net rate of heat transfer equation, and the emissivity can be found by rearranging the equation. The plate is diffuse-gray if its emissivity is between 0 and 1.
Explanation:The irradiation and emissivity of the plate can be found using the Stefan-Boltzmann law of radiation. The rate of heat transfer by emitted radiation is given by P = 6 AeT^4, where o = 5.67 × 10^-8 J/s. m². K^4 is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its temperature in kelvins.
To find the irradiation, we can use the equation Qnet = σe A (T₂^4 – T₁^4), where Qnet is the net rate of heat transfer, σ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, T₂ is the temperature of the surrounding air, and T₁ is the temperature of the plate.
To find the emissivity of the plate, we can rearrange the equation Qnet = σe A (T₂^4 – T₁^4) to solve for e. The plate is diffuse-gray if its emissivity is between 0 and 1. If the emissivity is 0, the plate is a perfect reflector, and if the emissivity is 1, the plate is a perfect absorber and emitter of radiation.
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The nucleus of a plutonium-239 atom contains 94 protons. Assume that the nucleus is a sphere with radius 6.64 fm and with the charge of the protons uniformly spread through the sphere. At the surface of the nucleus, what are the (a) magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons
Answer:
(a) E = 3.065 * 10^21 N/C
(b) The direction is radially outward.
Explanation:
Parameters given:
Number of protons = 94
Charge of proton = 1.6023 * 10^-19 C
Radius of nucleus = 6.65 fm = 6.65 * 10^-15
At the surface of the nucleus, Electric field, E is given as:
E = kq/r^2
Where k = Coulombs constant
r = radius of nucleus
=> E = (9 * 10^9 * 94* 1.6023 * 10^-19)/(6.65 * 10^21)
E = 3.065 * 10^21 N/C
The direction of the field is radially outward.
The magnitude of the electric field at the surface of a plutonium-239 nucleus is roughly 7.94 x 10^21 NC^-1. The direction of the electric field, created by positively charged protons, is radially outward.
Explanation:The question pertains to the electric field created by the protons of a plutonium-239 atom. The electric field resulting from a spherical charge distribution can be determined by using Coulomb's law transformed to a volumetric charge distribution. The formula is E = kQ/r2, where k is the Coulomb constant (8.99 x 109 Nm2/C2), Q is the total charge, and r is the distance from the center of the field (which in this case is the radius of the nucleus).
(a) Magnitude of the electric field: Here, the total charge Q is the charge of a single proton (1.60 x 10-19 C) multiplied by the total number of protons which is 94. Therefore, upon inserting these values into the formula, the calculated electric field at the surface of the nucleus is around 7.94 x 1021 NC-1.
(b) Direction of the electric field: The direction of the electric field is always from positive to negative. Since protons are positively charged and they are creating the field, the electric field direction will be radially outward.
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The same couple moment of 160 lb⋅ftlb⋅ft is required to open the hatch door, independent of where the couple's forces are applied. If the forces are applied at points C and D, what force magnitude FFF is required to open the hatch door?
Answer:
F = 88.88[lb-f]
Explanation:
To solve this problem, we must first look at the attached image.
Another important data that is needed to solve the problem is the distance from the Centre O to points C and D. The distance is 0.9 [ft].
Then we do a sum of moments around Point O and it equals zero.
[tex]SM_{O} =0\\F*0.9+F*0.9-160=0\\2*F*0.9=160\\F=88.88[lb-f][/tex]
Therefrore the couple of forces needed are 88.88[lb-f]
Two identical loudspeakers separated by distance dd emit 200 Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on.
What are the three lowest possible values of d? Assume a sound speed of 340 m/s.
Answer:
The first possible value of d is 0.85 m
The second possible value of d is 2.55 m
The third possible value of d is 4.25 m
Explanation:
Given that,
Distance =d
Frequency of sound wave= 200 Hz
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda=\dfrac{v}{f}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{340}{200}[/tex]
[tex]\lambda=1.7\ m[/tex]
The separation between the speakers in the destructive interference is
[tex]\Delta x= d[/tex]
The equation for destructive interference
[tex]2\pi\times\dfrac{\Delta x}{\lambda}-\Delta\phi_{0}=(m+\dfrac{1}{2})2\pi[/tex]
The loudspeakers are in phase
So, [tex]\Delta\phi_{0}=0[/tex]
The equation for destructive interference is
[tex]2\pi\times\dfrac{d}{\lambda}=(m+\dfrac{1}{2})2\pi[/tex]....(I)
Here, m = 0,1,2,3.....
We need to calculate the first possible value of d
For, m = 0
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{1}}{1.7}=(0+\dfrac{1}{2})2\pi[/tex]
[tex]d_{1}=\dfrac{1.7}{2}[/tex]
[tex]d_{1}=0.85\ m[/tex]
We need to calculate the second possible value of d
For, m = 1
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{2}}{1.7}=(1+\dfrac{1}{2})2\pi[/tex]
[tex]d_{2}=\dfrac{1.7\times3}{2}[/tex]
[tex]d_{2}=2.55\ m[/tex]
We need to calculate the third possible value of d
For, m = 1
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{3}}{1.7}=(2+\dfrac{1}{2})2\pi[/tex]
[tex]d_{3}=\dfrac{1.7\times5}{2}[/tex]
[tex]d_{3}=4.25\ m[/tex]
Hence, The first possible value of d is 0.85 m
The second possible value of d is 2.55 m
The third possible value of d is 4.25 m
2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofAB. Determine the moment of this force about for the two casesshown at below.
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.